A) Pushkar Singh Dhami to be sworn in as Uttarakhand CM on Sunday
Two-time MLA Pushkar Singh Dhami was chosen by the BJP legislature party and will be sworn in as chief minister of Uttarakhand on Sunday. Dhami’s selection follows the resignation of his predecessor Tirath Singh Rawat yesterday. Rawat had submitted his resignation to Governor Baby Rani Maurya as he was caught in a legal bind over his continuation as chief minister, an office, which required him to be elected as member of the Uttarakhand Assembly within six months of having taken oath. This, in turn, was rendered impossible by a clause in the Representation of the People Act which disallows by-polls for vacant seats if the tenure of the state assembly expires within a year. Uttarakhand goes to polls in March 2022. Left with no option, Rawat submitted his resignation paving the way for a new man to be elected by the party to head the state. New Uttarakhand CM Pushkar Dhami flanked by former CM T.S. Rawat and State BJP chief Madan Kaushik after Mr. Dhami was elected leader of the State legislature party. Forty-five-year-old Dhami will be the state’s10th chief minister and is considered to be close to Defence Minister Rajnath Singh. He will be the third chief minister in four months to head the state, beleaguered by serious infighting in the ruling party. Dhami was the officer on Special Duty to the state’s second chief minister Bhagat Singh Koshiari a year after the formation of the new state. Uttarakhand goes to polls in March 2022 and at the time of writing this, it is learnt, he is unlikely to be the face of the state where most of the MLAs have decided to fight the elections under Prime Minister Narendra Modi. Watch this space for this fast-developing story.
B) Punjab’s power crisis
Power and the availability of power may decide who comes to power in Punjab which is the grip of an unprecedented power crisis amidst soring mercury. The State has been witnessing some pitched protests by farmers and common people against the shortage of power. People are complaining that power cuts in the state lasts for hours on stretch. The demand for power is close to 14,225 MW while the supply is at 12,800 MW, a shortfall, the state government is unable to bridge. As paddy transplantation is in full-swing, farmers are reportedly upset that they are not getting their eight-hour power supply. Consumers have taken to the streets as they sweat it out in the summer heat. With the state government having to take some tough measures like staggered power supply in order to divert power for agricultural crops, there is also a ban on air conditioners in government offices. With the state slated to go to polls next year, opposition party AAP with 16 MLAs has promised 300 units of free power to every household in Punjab if voted to power next year. On Saturday, AAP party supporters protested in front of the residence of Chief Minister Amarinder Singh forcing the Punjab police to disperse the crowds with water cannons.
C) Covid Watch: Numbers and Developments
The number of reported coronavirus cases from India stood at 3,05,26,959 with the death toll at 4,01,516. The Union Government has sent multi-disciplinary teams to Kerala, Arunachal Pradesh, Tripura, Manipur, Odisha and Chattisgarh on Saturday, following a high number of COVID cases being reported from the states. The teams will monitor the COVID-19 management protocol being followed in the states assessing testing, surveillance, availability of beds, and vaccination progress. On the progress of vaccination, V K Paul, chairman of the National Empowered Group on Vaccine, told journalists that the estimate of 216 crore vaccine doses to be administered by year-end was aspirational and based on an optimistic assessment on what the vaccine manufacturers had conveyed to the Government. Till date 34 crore people have been vaccinated with at least one dose since the drive began on January 16, the government said.
INTERNATIONAL NEWS
A) Investigation of Rafale deal
A French judge will investigate the controversial 2016 ₹59,000 crore sale of 36 Rafale fighter jets to Indian government. Back home, the Opposition Congress claimed it has been vindicated. The party was engaged in a Twitter battle with the BJP on the matter. The deal inked between the Indian government and French aircraft manufacturer Dassault, originally initiated by the Congress-led UPA government in 2012 and executed by the Narendra Modi-led BJP Government in 2016, has been mired in controversies following allegations of hidden commission and payoffs to middle-men who facilitated the deal. The Parquet National financier (PNF), the French judicial institution responsible for tracking down serious economic and financial crimes, was reportedly reluctant to investigate even as several French mainstream newspapers and websites reported that there were shortcomings in the deal. French website Mediapart had even accused the French Anti-corruption Agency of burying suspicions surrounding the September 2016 deal, as AFP reported today. In a series of exposes in 2018, The Hindu, had also reported how the deal was unfavourable to India and was more expensive on the Indian exchequer; the report also exposed several shortcomings in the defence deal prompting the government to accuse the paper of stealing official documents. Several media reports today spoke of how three months ago, Mediapart had claimed millions of euros of hidden commissions were given to a go-between who helped Dassault conclude the sale, of which some… could have been given as bribes to Indian officials. Dassault responded that no wrong-doing was flagged in the group’s audits. After the reports, France’s Sherpa NGO, which specialises in financial crime, filed an official complaint for corruption and influence peddling among other accusations, prompting an investigating magistrate to be designated to probe the deal. In this first complaint, the NGO had pointed out how Dassault chose as its Indian partner, Reliance Group, which has had no previous experience on defence aircrafts. Dassault had initially won a contract in 2012 when the Congress-led United Progressive Alliance was in power, to supply 126 jets to India, and had been negotiating with Indian aerospace company Hindustan Aeronautics Limited (HAL). By March 2015, those talks had almost reached a conclusion, according to Dassault. Readers might recall how back in January 2016, at the time of the negotiations, Reliance had financed a film co-produced by Julie Gayet, the partner of Francois Hollande, who was president at the time, forcing Hollande to clarify that there was no conflict of interest. Hollande added in good measure that France had no say in who Dassault’s Indian partner was. The former president has repeatedly said that it was the Indian government that had suggested businessman Anil Ambani’s firm. France’s Le Monde newspaper also revealed that France in 2015 cancelled a 143.7-million-euro tax adjustment targeting a French firm belonging to Reliance, at the time when the deal was being negotiated. As investigations kick in, the Union Government will have its task cut out in defending its decision.
B) Twitter in crosshairs
Micro-blogging site Twitter caught in the crosshairs of the Union Government filed an affidavit in the High Court today in which it has claimed that it was in the last stages of hiring a grievance officer. Twitter had initially appointed Dharmendra Chatur, Principal Associate at Poovayya & Co., a Bangalore based law firm, who resigned from the post within a month prompting the micro-blogging site to appoint its California-based global legal policy director Jeremy Kessel as the new grievance officer for India. The new IT rules require an Indian resident for the job. The Union government is party to the petition in front of the court. The affidavit was in response to a notice issued by the Delhi High Court Twitter’s alleged non-compliance with the new information technology rules. Twitter and the Government of India have been engaged in a bitter feud with the latter accusing the California based firm of violating IT rules which mandate the appointment of a grievance official, an Indian resident based in India, to address the complaints of users. The new rules came into effect from May 25. IT Minister Ravi Shankar Prasad today praised social media platforms Google, Facebook and Instagram for publishing their first compliance report on voluntary removal of offensive posts as per new IT rules, terming it a big step towards transparency.
A) Trinamool seeks Tushar Mehta’s removal as Solicitor General over his meeting with BJP MLA Suvendu Adhikari; Mehta denies meeting him
The Trinamool Congress (TMC) on Friday wrote to Prime Minister Narendra Modi, seeking the removal of Solicitor General of India Tushar Mehta over his meeting with West Bengal Leader of the Opposition Suvendu Adhikari in New Delhi. The letter, written by TMC MPs Derek O’Brien, Sukhendu Sekhar Roy and Mahua Moitra, stated that the meeting between Adhikari and the Solicitor General (SG) reeks of impropriety as the BJP MLA is an accused in the Narada and Saradha cases where investigations are underway. According to the letter, the meeting curiously took place subsequent to Adhikari’s meeting with Home Minister Amit Shah. The TMC MPs claimed that Adhikari was accused in various cases of cheating, bribery and illegal gratification related to the Narada and Saradha scam cases. The SG is appearing for the CBI in the Narada case in the Supreme Court and the high court, besides advising the investigation agency in the Saradha chit fund scam, they said. Leader of the Opposition in the Bengal Assembly Suvendu Adhikari interacts with the media in Kolkata on July 2, 2021 as party leaders hold posters of alleged victims of post-poll violence. The meeting between Adhikari and the solicitor general not only reeks of impropriety, there is a direct conflict of interest and also taints the position of the second highest law officer of the country, the Solicitor General, they added. To maintain the neutrality and integrity of the office of the Solicitor General of India, necessary steps be taken for the removal of Mehta from the post, the TMC MPs said in the letter to the prime minister. Meanwhile, Mehta has issued a statement denying that he met the BJP leader. He claimed that Adhikari had come to his residence unannounced. Mehta said, Suvendu Adhikari came to my residence/office yesterday, unannounced. As I was in a pre-scheduled meet, my staff asked him to wait. After my meeting, my staff informed him of my inability to meet him. He left without insisting to meet me. Hence, the question of my meeting with him did not arise. Reacting to Mehta’s denial, Moitra tweeted, Z category BJP protectee enters high security official residence of Solicitor General of India uninvited, waits 20 mins over a cup of tea & apparently leaves without a mtng. CCTV footage to back your version, Mr. SG? Truth never been your forte!
B) Jammu drone attack an act of terror, says Air Chief Marshal Bhadauria; LeT could be behind it, says J&K DGP
Security agencies suspected the role of the Lashkar-e-Taiba (LeT) in the twin drone-driven blasts on the Indian Air Force’s technical airport in Jammu on June 27. The drones could have come from across the border, Jammu and Kashmir Director General of Police (DGP) Dilbagh Singh said on Friday. The investigation [into the airbase blasts] is in progress. We haven’t reached any conclusion as to who is actually involved. However, on the basis of the past history of the LeT using drones to drop weapons, narcotics and ready-made improvised explosive devices (IED), we suspect the outfit is involved, Singh said in Jammu. Meanwhile, Air Chief Marshal RKS Bhadauria said on Friday that the drone strikes were an act of terror aimed at targeting key military assets. He added that the Indian Air Force is in the process of bolstering its capabilities to deal with such security challenges. The Chief of Air Staff said the IAF has carried out a detailed analysis in terms of implications of drones and other similar capabilities falling into the hands of non-state actors, and taking a series of measures to counter them. What happened at Jammu was essentially a terrorist act which attempted to target our assets there. The attempt failed of course. The assets were not damaged. Two explosives were used, he said, in an interactive session at a think tank. Explosives-laden drones were used to carry out an attack on the Jammu Air Force station on Sunday in the first such instance of suspected Pakistan-based terrorists deploying unmanned aerial vehicles to strike at vital military installations in India. The Chief of Air Staff said a detailed investigation into the attacks was underway and that all sets of measures would be on the table based on the findings of the probe. We have gone over the subject in terms of the implications of this kind of capabilities in the hands of non-state actors and the kind of effect the armed drone capabilities would have in future conflicts, Air Chief Marshal Bhadauria said. He said the IAF analysed to a large extent the requirements in terms of planning and required systems and infrastructure to enhance its abilities to go for a soft kill and to have a counter-drone system.
C) Former Haryana CM Om Prakash Chautala released from Tihar Jail; plans to visit protesting farmers
Released from Tihar Central Jail on Friday on completion of a 10-year jail term in a teachers’ recruitment scam, former Haryana Chief Minister and Indian National Lok Dal (INLD) leader Om Prakash Chautala said he planned to visit farmers at protest sites. Addressing party workers assembled at the Delhi-Gurugram border to welcome the octogenarian leader, Chautala thanked them for supporting his party against the wrong policies of the government and assured them that he would continue to fight for the poor, the downtrodden and the farmers. Chautala said doctors had advised him a few days of rest, after which he would visit the protesting sites to meet farmers and seek their blessings. He said he will visit villages across the country. He was jailed in 2013 in connection with the Junior Basic Training teachers recruitment scam case. He was on emergency parole since March 26, 2020, due to the Covid-19 pandemic. Sunaina Chautala, principal general secretary, women’s cell, INLD, said the release of Chautala had infused new energy into the workers and the party would return to power with a thumping majority. A large number of workers holding party flags with the INLD symbol and sporting green headgear turned up at the Delhi-Gurugram border on National Highway-48 to welcome Chautala.
D) Assets of Ahmed Patel’s son-in-law, Dino Morea, DJ Aqeel attached in money-laundering case
The Enforcement Directorate on Friday said it has attached the assets of late Congress leader Ahmed Patel’s son-in-law, actors Dino Morea and Sanjay Khan, and DJ Aqeel in a money-laundering case involving Gujarat-based pharmaceutical company Sterling Biotech group, PTI reported. It said a provisional order was issued under the Prevention of Money Laundering Act (PMLA) to attach the assets of the four people, valued at ₹8.79 crore. Out of this, the value of attachment of assets for Khan is ₹3 crore, for Dino Morea it is ₹1.4 crore, for Aqeel Abdulkhalil Bachooali, popularly known as DJ Aqeel, it is ₹ 1.98 crore, and for Irfan Ahmed Siddiqui, who is Patel’s son-in-law, it is ₹2.41 crore, the central probe agency said in a statement. The Enforcement Directorate (ED) said that Nitin Sandesara and Chetan Sandesara, absconding promoters of the Sterling Biotech group, have diverted proceeds of their crime to the four people. The promoter brothers Nitin Sandesara, Chetan Sandesara, Chetan’s wife Dipti Sandesara and Hitesh Patel have been declared fugitive economic offenders by a special court, it said. The money-laundering case pertains to the alleged ₹14,500-crore bank-loan fraud said to have been perpetrated by Sterling Biotech and its main promoters and directors.
E) Police action over actor Chetan Kumar’s remarks on ‘brahminism’ sparks row in Karnataka
Kannada actor and activist Chetan Kumar has been questioned thrice this month after an FIR was filed by the Bengaluru Police for his critical remarks on brahmanism. Some members of the Brahmin community have objected to his comments, arguing that the very use of the word was defamatory and abusive and had hurt their sentiments. Attempts to criminalise use of the term brahmanism — widely used by several anti-caste thinkers including Dr. B. R. Ambedkar and Periyar E. V. Ramasamy to denote caste-based hierarchy, discrimination and notions of superiority — have sparked a controversy in Karnataka. While several writers and intellectuals have rallied behind the actor, his detractors have been vociferous, especially on social media. State Labour Minister A. Shivaram Hebbar, also a Brahmin, termed the use of the term defamatory and called for Kumar’s arrest. He also alleged the actor was making these statements for monetary gains, prompting the latter to sue the Minister for defamation seeking damages of ₹1 and a public apology. Defending his remarks, Kumar said he had only rearticulated what Bahujan thinkers have said.
F) Deciding cases not an easy task, have to keep in mind repercussions of setting a precedent, says CJI N.V. Ramana
Judging is not an easy task, Chief Justice of India (CJI) N.V. Ramana said on Friday. On one hand, the judge should focus on law, the precedents and facts of a case and, on the other, the human aspect. Judges should use the little discretion law allows to keep sight of the human suffering and toll behind every case. A decision of the court would echo through time. It would have repercussions. A judgment becomes the law of the land. A judge had to keep all this in mind while dealing with a case, he stated. Deciding cases is not an easy task. We not only have to focus on the law and precedents surrounding the issue before us, as well as the facts of the case, but also the repercussions of what we decide and the precedent we may be setting. This makes it necessary for us judges to be logical and objective and theoretically sound. However, we should not lose sight of the people and their difficulties behind the cases. The little discretion that is given to us, is the area in which a judge has flexibility to display his philosophy, he observed. The CJI was speaking at a farewell organised by the Supreme Court Bar Association for retiring judge Justice Ashok Bhushan. He said Justice Bhushan, besides his remarkable judgments, had, as a humanist judge, left a mark in the hearts and minds of the people of the country. The CJI termed the Bar the protector of the institution. He said, Lawyers must respect the institution and protect the judiciary from any onslaught which is likely to affect the functioning of the judicial system. I know, the Supreme Court Bar is always a frontrunner when it comes to supporting the institution from motivated attacks. They always cherish the contribution of judges and appreciate their hard work. They never forget the efforts of judges. The CJI underscored that the strength of the institution lay in the unity of the Bar and the Bench.
INTERNATIONAL NEWS
A) Six million EU citizens apply to stay in U.K.
More than six million EU citizens have applied to stay in the U.K. after Brexit, the government said on Friday, nearly twice the number thought be living in Britain before it left the bloc. There was a late flurry of applications ahead of a June 30 deadline for the settlement scheme, designed to allow EU citizens residing in the U.K. to retain the same rights as they enjoyed before Brexit. But since the end of the Brexit transition period on January 1, EU citizens wishing to settle or stay have faced tougher tests to secure residency and employment. The final figure of 6.02 million applications, with 5.1 million grants of status, is far higher than the 3.7 million EU nationals originally estimated to be in the country when the scheme launched in March 2019. More than 4,00,000 applications were made in the final month before the scheme ended, the Home Office said, adding that those who had missed the deadline on reasonable grounds could still make a late application. When we left the EU we promised to protect the rights of EU citizens who have made their life in the U.K ., and developed the hugely successful EU Settlement Scheme to ensure they could call the U.K. home in the years to come, Home Secretary Priti Patel said. The government has touted the settlement scheme as the successful delivery of promises made to EU nationals in the U.K. after Britain’s withdrawal from the bloc in December and the end of free movement. However, immigration rights campaigners have complained that the most vulnerable – including children placed in care, young adults previously in care, the homeless and the elderly – have been allowed to slip through the scheme’s net.
B) U.S. issues moratorium on death penalty at federal level.
The U.S. Department of Justice (DoJ) has issued a moratorium on federal executions while it reviews policies and procedures, the Department said in a statement. U.S. President Joe Biden had said on his campaign website that he would legislate the end of capital punishment at the federal level and incentivise states to follow suit. The Department of Justice must ensure that everyone in the federal criminal justice system is not only afforded the rights guaranteed by the Constitution and laws of the United States, but is also treated fairly and humanely. That obligation has special force in capital cases. Attorney General Merrick Garland said in a statement released on Thursday. Apart from the moral case against the death penalty, the data show that in its application, it is biased against racial minorities, especially African Americans. Under former President Donald Trump, the federal government carried out 13 executions. This is the highest number of executions under any presidency since the 19th Century, Reuters reported. The review, ordered by Mr. Garland, will include an assessment of the risk of pain and suffering caused by pentobarbital – a lethal injection drug. The Trump administration’s DoJ had adopted a single drug, instead of a three drug protocol. The review will also look into a November 2020 expansion of federal execution methods beyond lethal injection and policy changes to expedite executions.
Q. 1: An electron with kinetic energy K1 enters between parallel plates of a capacitor at an angle ‘α’ with the plates. It leaves the plates at angle ‘β’ with kinetic energy K2. Then the ratio of kinetic energies K1 : K2 will be:
(A) sin2β/cos2α
(B) cos2β/cos2α
(C) cosβ/sinα
(D) cosβ/cosα
Answer: (B) ∵v1cosα = v2cosβ [electric field inside a parallel plate capacitor is perpendicular to the plates, hence, there will be no change in parallel component of velocity] v1/v2 = cosβ/cosα Then the ratio of kinetic energies k1/k2 = ½ mv12/ ½ mv22 = (v1/v2)2 = (cosβ/cosα)2 k1/k2 = cos2β/cos2α
Q. 2: Two identical springs of spring constant ‘2K’ are attached to a block of mass m and to fixed support (see figure). When the mass is displaced from equilibrium position on either side, it executes simple harmonic motion. Then, time period of oscillations of this system is:
(A) π√(m/k)
(B) π√(m/2k)
(C) 2π√(m/k)
(D) 2π√(m/2k)
Answer: (A) Springs are in parallel combination. Hence, Keff = 2k + 2k = 4k ∵ T = 2π √(m/keff) = 2π√(m/4k) T = π √(m/k)
Q. 3: The wavelength of the photon emitted by a hydrogen atom when an electron makes a transition from n = 2 to n = 1 state is:
(A) 194.8 nm
(B) 490.7 nm
(C) 913.3 nm
(D) 121.8 nm
Answer: (D) ∆E = 10.2 eV is the energy difference between the state n = 2 & n = 1 ∆E = –3.4 – (–13.6) = 10.2 eV hc/λ = 10.2 ev λ = hc/(10.2)e (in meters) where ‘e’ = 1.6 × 10–19 J/V = 12400/10.2 Å (because hc = 12400 eV nm) = 121.56 nm ≃ 121.8 nm
Q. 4: In a ferromagnetic material below the Curie temperature, a domain is defined as:
(A) a macroscopic region with consecutive magnetic dipoles oriented in opposite directions.
(B) a macroscopic region with zero magnetization.
(C) a macroscopic region with saturation magnetization.
(D) a macroscopic region with randomly oriented magnetic dipoles.
Answer: (C) In a ferromagnetic material below the Curie temperature, the domain is defined as a macroscopic region with saturation magnetization.
Q. 5: The point A moves with a uniform speed along the circumference of a circle of radius 0.36m and covers 30o in 0.1s. The perpendicular projection ‘P’ form ‘A’ on the diameter MN represents the simple harmonic motion of ‘P’. The restoring force per unit mass when P touches M will be:
(A) 100 N
(B) 50 N
(C) 9.87 N
(D) 0.49 N
Answer: (C) The point covers 30o in 0.1 sec. Means π/6 → 0.1sec. 1 → 0.1/(π/6) 2π = → 0.1×6×2π/π T = 1.2 sec. We know that ω = 2π/T ω = 2π/1.2 Restoring force (F) = mω2A Then, Restoring force per unit mass (F/m) = ω2A F/m = (2π/1.2)2×0.36 ≃ 9.87 N
Q. 6: Match List I with List II.
List I
List II
(a) Rectifier
(i) Used either for stepping up or stepping down the A.C.voltage
(b) Stabilizer
(ii) Used to convert A.C. voltage into D.C. voltage
(c) Transformer
(iii) Used to remove any ripple in the rectified output voltage
(d) Filter
(iv) Used for constant output voltage even when the input voltage or load current change
Choose the correct answer form the options given below:
(A) (a)-(ii), (b)- (i), (c)-(iv), (d)-(iii)
(B) (a)-(ii), (b)- (iv), (c)-(i), (d)-(iii)
(C) (a)-(ii), (b)- (i), (c)-(iii), (d)-(iv)
(D) (a)-(iii), (b)- (iv), (c)-(i), (d)-(ii)
Answer: (B) (a) Rectifier:- used to convert A.C voltage into D.C. Voltage. (b) Stabilizer:- used for constant output voltage even when the input voltage or load current change (c) Transformer:- used either for stepping up or stepping down the A.C. voltage. (d) Filter:- used to remove any ripple in the rectified output voltage.
Q. 7: Y = A sin(ωt + ϕ) is the time – displacement equation of an SHM, At t = 0, the displacement of the particle is Y = A/2 and it is moving along negative x-direction. Then, the initial phase angle ϕ will be.
(A) π/6
(B) π/3
(C) 2π/3
(D) 5π/6
Answer: (D) y = A sin (ωt + ϕ) t = 0, x = A/2 1/2 = sin ϕ ϕ = π/6, 5π/6 v = dy/dt = A ωcos(ωt + ϕ) t = 0, v = A ωcosϕ ϕ = π/6, for v (positive) ϕ = 5π/6, for v (negative) ∴ϕ = 5π/6
Q. 8: A sphere of radius ‘a’ and mass ‘m’ rolls along horizontal plane with constant speed v0. It encounters an inclined plane at angle θ and climbs upwar(D) Assuming that it rolls without slipping how far up the sphere will travel (along the incline)?
(A) (⅖) v02/g sin θ
(B) 10v02/7gsin θ
(C) v02/5gsin θ
(D) 7v02/10gsin θ
Answer: (D) From energy conservation mgh = ½ mv2 + ½ Iω2 mgh = ½ mv2 + ½×(⅖)ma2× v02/a2 gh = ½ v02+ ⅕ v02 gh = (7/10)v02 h = (7/10)v02/g From triangle, sinθ = h/l Then, h = lsinθ l sinθ = (7/10) (v02/g) l = (7/10) (v02/g sin θ)
Q. 9: Consider the diffraction pattern obtained from the sunlight incident on a pinhole of diameter 0.1 μm. If the diameter of the pinhole is slightly increased, it will affect the diffraction pattern such that:
(A) its size decreases, but intensity increases
(B) its size increases, but intensity decreases
(C) its size increases, and intensity increases
(D) its size decreases, and intensity decreases
Answer: (A) For diffraction through a single slit, for first minimum. sin θ = 1.22λ/D If D is increased, then sinθ will decrease i.e θ will decrease ∴ size of circular fringe will decrease but intensity increases.
Q. 10: An electron of mass me and a proton of mass mp = 1836 me are moving with the same speed. The ratio of their de Broglie wavelength λelectron/λProton will be:
(A) 918
(B) 1836
(C) 1/1836
(D) 1
Answer: (B) Given mass of electron = me Mass of proton = mp ∴ given mp = 1836 me From de-Broglie wavelength λ = h/p = h/mv λe/λp = mp/me [v is same] = 1836me/me λe/λp= 1836
Q. 11: The stopping potential for electrons emitted from a photosensitive surface illuminated by light of wavelength 491 nm is 0.710 V. When the incident wavelength is changed to a new value, the stopping potential is 1.43 V. The new wavelength is:
(A) 400 nm
(B) 382 nm
(C) 309 nm
(D) 329 nm
Answer: (B) From the photoelectric effect equation hc/λ = ϕ + evs where Vs is the stopping potential and ϕ is work function of the metal. So, evs1 = hc/λ1 – ϕ ……(i) evs2 = hc/λ2 – ϕ …..(ii) Subtract equation (i) from equation (ii) evs1 – evs2 = hc/λ1– hc/λ2 vs1 – vs2 = (hc/e) (1/λ1 – 1/λ2) (0.710 – 1.43) = 1240 (1/491 – 1/λ2) (because hc = 1240 eV nm) -0.72/1240 = 1/491 – 1/λ2 1/λ2 = 1/491 + 0.72/1240 1/λ2 = 0.00203 + 0.00058 1/λ2 = 0.00261 λ2 = 383.14 λ2 ≃ 382 nm
Q. 12: The truth table for the following logic circuit is:
Answer: (D) If A = B = 0 then output y = 1 If A = B = 1 then output y = 1
Q. 13: If e is the electronic charged, c is the speed of light in free space and h is planck’s constant, the quantity (1/4πε0) |e|2/hc has dimensions of:
(A) [ LC-1]
(B) [M0 L0 T0]
(C) [ M L T0]
(D) [M L T-1]
Answer: (B) Given e = electronic charge c = speed of light in free space h = planck’s constant (1/4πε0) e2/hc = (ke2/hc)× λ2/λ2 [multiply and divide by lambda = wavelength] = F×λ/E [ke2/λ2 has dimensions of force and hc/λ = E] = E/E [F× λ has dimension of E] = dimensionless = [ M0 L0 T0]
Q. 14: A charge ‘q’ is placed at one corner of a cube as shown in figure. The flux of electrostatic field E→through the shaded area is:
(A) q/48ε0
(B) q/8ε0
(C) q/24ε0
(D) q/4ε0
Answer: (C) Total flux through the cube = (q/ε0)× ⅛ = q/8ε0 Total flux through one “outer” face of the cube = (q/8ε0)×1/3 = q/24ε0 [Because there is flux only through 3 faces] Hence, total flux through shaded area ϕT = (q/24ε0 + q/24ε0)×1/2 [half of each face is shaded] ϕT = q/24ε0
Q. 15: Thermodynamic process is shown below on a P-V diagram for one mole of an ideal gas. If V2 = 2V1 then the ratio of temperature T2/T1 is:
(A) 1/√2
(B) 1/2
(C) 2
(D) √2
Answer: (D) From P-V diagram, Given PV1/2 = constant …..(i) We know that PV = nRT P ∝ (T/V) [for 1 mole] Put in equation (i) (T/V) (V)1/2 = constant T ∝V1/2 T2/T1 = √(V2/V1) T2/T1 = √(2V1/V1) T2/T1 = √2
Q. 16: Given below are two statements:
Statement I: In a diatomic molecule, the rotational energy at a given temperature obeys Maxwell’s distribution.
Statement II: In a diatomic molecule, the rotational energy at a given temperature equals the transnational kinetic energy for each molecule.
In the light of the above statements, choose the correct answer from the options given below:
(A) Both statement I and statement II are false.
(B) Both statement I and statement II are true.
(C) Statement I is false but statement II is true.
(D) Statement I is true but statement II is false.
Answer: (D) The translational kinetic energy & rotational kinetic energy both obey Maxwell’s distribution independent of each other. T.K.E. of diatomic molecules = (3/2) kT [3 translational D.O.F.] R.K.E. of diatomic molecules = (2/2) kT = [2 rotational D.O.F.] So statement I is true but statement II is false.
Q. 17: An LCR circuit contains resistance of 110 Ω and a supply of 220 V at 300 rad/s angular frequency. If only capacitance is removed from the circuit, current lags behind the voltage by 450. If on the other hand, only inductor is removed the current leads by 450 with the applied voltage. The rms current flowing in the circuit will be:
(A) 2.5 A
(B) 2 A
(C) 1 A
(D) 1.5 A
Answer: (B) When L and C are connected with R in series the circuit will come in resonance So, current in the circuit will be: Irms = Vrms/R = 220/110 = 2 A
Q. 18: For extrinsic semiconductors: when doping level is increased;
(A) Fermi–level of p and n-type semiconductors will not be affected.
(B) Fermi–level of p-type semiconductors will go downward and Fermi–level of n-type semiconductor will go upward.
(C) Fermi–level of both p–type and n–type semiconductors will go upward for T >TFK and downward for T<TFK, where TF is Fermi temperature.
(D) Fermi–level of p-type semiconductor will go upward and Fermi–level of n–type semiconductors will go downward.
Answer: (B) In n-type semiconductor, pentavalent impurity is added Each pentavalent impurity donates a free electron. So the Fermi-level of n-type semiconductor will go upward and in p-type semiconductor, trivalent impurity is added. Each trivalent impurity creates a hole in the valence band So the Fermi-level of p-type semiconductor will go downward.
Q. 19: A stone is dropped from the top of a building. When it crosses a point 5m below the top, another stone starts to fall from a point 25m below the top, both stones reach the bottom of the building simultaneously. The height of the building is: [Take g = 10 m/s2]
(A) 45 m
(B) 35 m
(C) 25 m
(D) 50 m
Answer: (A) Velocity of particle (1) at 5m below top u1 = √2gh = √(2×10×5) =10 m/s For particle (1), using 2nd equation of motion 20+h = 10t + ½ gt2 …..(i) For particle (2), using 2nd equation of motion h = ½ gt2 …..(ii) Put equation (ii) in equation (i) 20 + ½ gt2 = 10t + ½ gt2 t = 2 sec. Put in equation (ii) h = ½ gt2 = ½ ×10 × 22 h = 20m The height of the building = 25 + 20 = 45m
Q. 20: If a message signal of frequency ‘fm’ is amplitude modulated with a carrier signal of frequency ‘fc’ and radiated through an antenna, the wavelength of the corresponding signal in air is:
[Given, C is the speed of electromagnetic waves in vacuum/air]
(A) c/(fc+fm)
(B) c/(fc-fm)
(C) c/fm
(D) c/fc
Answer: (A) Equation of amplitude modulated wave y = (Ac + Am sin ωmt) sin ωct Here angular frequency of modulated signal = ωc Thus frequency of modulated signal = fc Thus wavelength = c/fc
Section B
Q. 1: The initial velocity v1 required to project a body vertically upward from the surface of the earth to just reach a height of 10R, where R is the radius of the earth, may be described in terms of escape velocity ve such that vi = √(x/y) ve. The value of x will be_______.
Answer: 20 Here R = radius of the earth From energy conservation -Gmem/R + ½ mvi2 = -Gmem/11R + 0 ½ mvi2 = (10/11)Gmem/R vi = √(20Gme/11R) vi = √(20/11) ve {∵escape velocity ve = √(Gme/R)} Then the value of x = 20
Q. 2: The percentage increase in the speed of transverse waves produced in a stretched string if the tension is increased by 4% will be ___%.
Answer: 2 Speed of transverse wave is V = √(T/μ) Taking log on both sides ln v = ½ lnT – ½ ln μ Δv/v = ½ ΔT/T ⇒ (Δv/v)×100 = 1/2 ( ΔT/T)×100 = ½ ×4 [ μ is constant for a string] (Δv/v)×100 = 2%
Q. 3: The peak electric field produced by the radiation coming from the 8 W bulb at a distance of 10 m is (x/10) √(µ0c/π) V/m. The efficiency of the bulb is 10% and it is a point source. The value of x is ___.
Answer: 2 Intensity I = ½ c∈0E02 Intensity = Power / Area = 8 /(4π × 102) (8/4π×102) ×½ =¼ × c ×(1/µ0c2) ×E02 [Multiply by ½ and ∈0 =1/μ0c2] E0 = (2/10)√(µ0c/π) ⇒ x = 2
Q. 4: Two identical conducting spheres with negligible volume have 2.1nC and -0.1nC charges, respectively. They are brought into contact and then separated by a distance of 0.5 m. The electrostatic force acting between the spheres is _______× 10-9 N. [Given : 4πε0 = 1/9×109 SI unit]
Answer: 36 When they are brought into contact & then separated by a distance = 0.5 m. Then charge distribution will be The electrostatic force acting b/w the sphere is Fe = kq1q2/r2 = 9×109×1×10-9×1×10-9/(0.5)2 = (900/25) × 10-9 Fe = 36 ×10-9 N
Q. 5: A current of 6A enters one corner P of an equilateral triangle PQR having 3 wires of resistance 21Ω each and leaves by the corner R. The currents i1 in ampere is______.
Answer: 2 The current i1 = (R2/(R1 + R2))i = (2/(4+2))×6 i1 = 2A
Q. 6: The wavelength of an X-ray beam is 10 Å. The mass of a fictitious particle having the same energy as that of the X-ray photons is (x/3) h kg. The value of x is _______.
Answer: 10 Given wavelength of an X-ray beam = 10 Å ∵E = hc/λ = mc2 m = hc/λ will be the mass of a particle having the same energy The mass of a fictitious particle having the same energy as that of the X-ray photons = (x/3)h kg (x/3) h = h/cλ x = 3/cλ = 3/3×108×10×10-10 x = 10
Q. 7: A reversible heat engine converts one- fourth of the heat input into work. When the temperature of the sink is reduced by 52 K, its efficiency is doubled The temperature in Kelvin of the source will be_______.
Answer: 208 ∵ n = W/Qin = 1/4 ¼ = 1 – T2/T1 [for reversible heat engine] T2/T1 = 3/4 When the temperature of the sink is reduced by 52k then its efficiency is doubled. 1/2 = 1 -(T2 – 52)/T1 (T2– 52)/T1 = 1/2 T2/T1 – 52/T1 = 1/2 ¾ – 52/T1 = 1/2 52/T1 = 1/4 T1 = 208 K is the source temperature.
Q. 8: If P￫ × Q￫ = Q￫ × P￫, the angle between P￫ and Q￫ is θ(0o<θ< 360o). The value of ‘θ’ will be ____.
Answer: 180
Q. 9: Two particles having masses 4g and 16g respectively are moving with equal kinetic energies. The ratio of the magnitudes of their linear momentum is n:2. The value of n will be_____.
Answer: 1 Relation b/w kinetic energy & momentum is p = √(2mKE) (∵KE = same) p1/p2 = √(m1/m2) n/2 = √(4/16) n = 1
Q. 10 : Two small spheres each of mass 10 mg are suspended from a point by threads 0.5 m long. They are equally charged and repel each other to a distance of 0.20 m. Then charge on each of the sphere is (a/21)×10-8C. The value of ‘a’ will be________. [Take g = 10 m/s2]
Answer: 20 From FBD, T sinθ = kq2/d2 and T cosθ = mg Dividing them, mg tanθ = kq2/d2 q = √(mg tanθ)/25k = √(10×10-6×10×1/√24(25)9×109) = √(10-4×4/√24×25×4×9×109) = (⅔)√(10-4/√24×1011) Thus (a/21)×10-8 = ⅔ √(10-15/√24) = ⅔ √(10-16/0.49) a = 2×21/3×0.7 = 20
CHEMISTRY
SECTION A
Q. 1. What is ‘X’ in the given reaction?
Answer: (A)
Q. 2. The major product of the following reaction:
Answer: (D)
Q. 3. Given below are two statements:
Statement I: The identification of Ni2+ is carried out by dimethylglyoxime in the presence of NH4OH
Statement II: The dimethylglyoxime is a bidentate neutral ligand.
In the light of the above statements, choose the correct answer from the options given below:
(A) Both statement I and statement II are true
(B) Both statement I and statement II are false
(C) Statement I is false but statement II is true
(D) Statement I is true but statement II is false
Answer: (D) Dimethylglyoxime is a negative bidentate legend.
Q. 4. Carbylamine test is used to detect the presence of a primary amino group in an organic compound Which of the following compounds is formed when this test is performed with aniline?
Answer: (B)
Q. 5. The correct order of bond dissociation enthalpy of halogen is:
(A) F2>Cl2>Br2>I2
(B) Cl2>F2>Br2>I2
(C) Cl2>Br2>F2>I2
(D) I2>Br2>Cl2>F2
Answer: (C) Fact based F2 has F — F, F2 involves repulsion of non-bonding electrons & more over its size is small and hence due to high repulsion its bond dissociation energy is very low.
Q. 6. Which one of the following statements is FALSE for hydrophilic sols?
(A) These sols are reversible in nature
(B) The sols cannot be easily coagulated
(C) They do not require electrolytes for stability
(D) Their viscosity is of the order of that of H2O
Answer: (D) Fact based
Q. 7. Water does not produce CO on reacting with
(A) C3H8
(B) C
(C) CH4
(D) CO2
Answer: (D) H2O + CO2 H2CO3
Q. 8: The correct sequence of reagents used in the preparation of 4-bromo-2-nitroethyl benzene from benzene is:
Answer: (D) Fact German silver is alloy which does not have silver. Cu-50%; Ni-30%; Zn-20%
Q. 15. The method used for the purification of Indium is:
(A) Van Arkel method
(B) Vapour phase refining
(C) Zone refining
(D) Liquation
Answer: (C) Fact Ga, In, Si, Ge are refined by zone refining or vacuum refining.
Q. 16. Which of the following is correct structure of α-anomer of maltose
Answer: (D)
Q. 17. The major product of the following reaction is:
(A) CH3CH2CH2CHO
(B) CH3CH2CH=CH–CHO
(C) CH3CH2CH2CH2CHO
(D)
Answer: (C)
Q. 18. The correct order of acid character of the following compounds is:
(A) II>III>IV>I
(B) III>II>I>IV
(C) IV>III>II>I
(D) I>II>III>IV
Answer: (A) Acidity of carboxylic acid α-R>-H>-I 1 / α + R > + H > + I
Q. 19. Which among the following species has unequal bond lengths?
(A) XeF4
(B) SiF4
(C) BF4-
(D) SF4
Answer: (D)
Q. 20. If which of the following orders the given complex ions are arranged correctly with respect to their decreasing spin only magnetic moment?
(i) [FeF6]3-
(ii) [Co(NH3)6]3+
(iii) [NiCl4]2-
(iv) [Cu(NH3)4]2+
(A) (ii)>(i)>(iii)>(iv)
(B) (iii)>(iv)>(ii)>(i)
(C) (ii)>(iii)>(i)>(iv)
(D) (i)>(iii)>(iv)>(ii)
Answer: (D) [FeF6]3-: Fe3+ 3d5 → 5-unpaired electrons as F- is weak field legend [Co(NH3)6]3+: Co3+ 3d6→ No-unpaired electron as NH3 is strong field light and causes pairing [NiCl4]2-: Ni2+ 3d8 → 2-unpaired electrons [Cu(NH3)4]2+: Cu2+ 3d9 → 1-unpaired electrons
SECTION B
Q. 1. Consider titration of NaOH solution versus 1.25 M oxalic acid solution. At the end point following burette readings were obtained.
(i) 4.5 mL (ii) 4.5 mL (iii) 4.4 mL (iv) 4.4 mL (v) 4.4 mL
If the volume of oxalic acid taken was 10.0 ml. then the molarity of the NaOH solution is ____M. (Rounded-off to the nearest integer)
Q. 2. Electromagnetic radiation of wavelength 663 nm is just sufficient to ionize the atom of metal A. The ionization energy of metal A in kJ mol—1 is____. (Rounded off to the nearest integer)
[h=6.63×10-34Js, c = 3.00×108ms-1, NA=6.02×1023 mol-1]
Answer: 180 Energy required to ionize an atom of metal ‘A’ = hc / λ = hc / 663 nm for 1 mole atoms of ‘A’ Total energy required = NA × hc / λ =
Q. 3. Copper reduces into NO and NO2 depending upon the concentration of HNO3 in solution. (Assuming fixed [Cu2+] and PNO=PNO2), the HNO3 concentration at which the thermodynamic tendency for reduction of into NO and NO2 by copper is same is 10x M.
The value of 2x is ______. (Rounded-off to the nearest integer)
[Given:
E0(Cu2+ /Cu) = 0.34V, E0 (No3– / NO) = 0.96V, E0(No3– / NO2 = 0.79V and at 298 K,
Q. 4. Five moles of an ideal gas at 293 K is expanded isothermally from an initial pressure of 2.1 MPa to 1.3 MPa against a constant external 4.3 MP(A) The heat transferred in this process is ____kJ mol-1. (Rounded-off of the nearest integer)
[Use R = 8.314 J mol-1 K-1]
Answer: 15 Moles (n) = 5 T = 293 K Process = Iso T. → Irreversible Pini = 2.1 M Pa Pt = 1.3 M Pa Pext = 4.3 mPa Work = – Pext Δv
=−4.3×(5×293R / 1.3 − 5×293/2.1)
=−5×293×8.314×43(1/13 −1/21)
=(5×293×8.314×43×8)/(21×13)
= -15347.7049 J = – 15.34 kJ Isothermal process, so ΔU = 0 w = – Q Q = 15.34 kJ / mol So, answer is 15.
Q. 5. Among the following, the number of metal/s which can be used as electrodes in the photoelectric cell is _____(Integer answer).
Answer: (A) Cs is used in photoelectric cells due to its very low ionization potential.
Q. 6. The rate constant of a reaction increases by five times on increasing temperature from 270 C to 520C. The value of activation energy in kJ mol-1 is ______.
(Rounded off to the nearest integer)
[R=8.314 J K-1 mol-1]
Answer: 52 Ea = 51524.96 J/mol Ea = 51.524 kJ/mol 52 Ans.
Q. 7. If a compound AB dissociates to the extent of 75% in an aqueous solution, the molality of the solution which shows a 2.5 K rise in the boiling point of the solution is ______molal.
Answer: 3 AB → A+ + B– 1 – α α α α = 3 / 4 N = 2 i = [1+(2-1)α] 2.5 = [1+(2-1)3/4] × 0.52 × m
Q. 8. The spin only magnetic moment of a divalent ion in aqueous solution (atomic number 29) is ____BM.
Answer: 2 29Cu+2 → [Ar]183d9 No. of unpaired e– = 1 Magnetic moment = μ = √n(n + 2) μ = √(1)(1 + 2) = √3B.M. = 1.73 B.M
Q. 9. The number of compound/s given below which contain/s —COOH group is ______
(A) Sulphanilic acid
(B) Picric acid
(C) Aspirin
(D) Ascorbic acid
Answer: (A)
Q. 10. The unit cell of copper corresponds to a face centered cube of edge length 3.596 Å with one copper atom at each lattice point. The calculated density of copper in kg/m3 is ______.
Q. 3: A hyperbola passes through the foci of the ellipse x2/25 + y2/16 = 1 and its transverse and conjugate axes coincide with major and minor axes of the ellipse, respectively. If the product of their eccentricities is one, then the equation of the hyperbola is:
(A) x2/9 – y2/4 = 1
(B) x2/9 – y2/16 = 1
(C) x2 – y2 = 9
(D) x2/9 – y2/25 = 1
Answer: (B) For ellipse, e1 = √(1–16/25)=3/5 Foci = (±3,0) Let equation of hyperbola be x2/A2 – y2/B2 = 1, passes through (±3, 0) A2 =9, A=3, e2=5/3 e22 = 1 + B2/A2 25/9 = 1 + B2/9 ⇒B2 = 16 x2/9 – y2/16 = 1
Q. 4: limn→∞[1/n + n/(n+1)2 + n/(n+2)2 + …….. + n/(2n-1)2] is equal to:
(A) 1
(B) 1/3
(C) 1/2
(D) 1/4
Answer: (C)
Q. 5: In a group of 400 people, 160 are smokers and non-vegetarian; 100 are smokers and vegetarians and the remaining 140 are non-smokers and vegetarian. Their chances of getting a particular chest disorder are 35%, 20% and 10% respectively. A person is chosen from the group at random and is found to be suffering from chest disorder. The probability that the selected person is a smoker and non-vegetarian is:
(A) 7/45
(B) 8/45
(C) 14/45
(D) 28/45
Answer: (D) Based on Bayes’ theorem Probability = ((160/400×35/100))/((160/400×35/100)+(100/400×20/100)+(140/400×10/100) ) = 5600/9000 = 28/45
Q. 6: The following system of linear equations
2x + 3y + 2z = 9
3x + 2y + 2z = 9
x – y + 4z = 8
(A) does not have any solution
(B) has a unique solution
(C) has a solution (α, β, γ) satisfying α + β2 + γ3 = 12
(D) has infinitely many solutions
Answer: (B) Determinant of given system = -20 ≠ 0 It has a unique solution.
Q. 7: The minimum value of f(x) = a^ax + a^(1-ax) where a, x ∈ R and a > 0, is equal to:
(A) a + 1/a
(B) a + 1
(C) 2a
(D) 2√a
Answer: (D) Using AM ≥ GM inequality, we get
Q. 8: A function f(x) is given by f(x) = 5x/(5x+5), then the sum of the series f(1/20) + f(2/20) + f(3/20) + …… + f(39/20) is equal to:
Q. 9: A plane passes through the points A(1,2,3), B(2,3,1) and C(2,4,2). If O is the origin and P is (2,–1,1), then the projection of vector(OP) on this plane is of length:
(A) √(2/5)
(B) √(2/3)
(C) √(2/11)
(D) √(2/7)
Answer: (C) A(1,2,3),B(2,3,1),C(2,4,2),O(0,0,0) Equation of plane passing through A, B, C will be (x – 1)(–1 + 4) – (y – 2)(–1 + 2) + (z – 3)(2 – 1) = 0 (x – 1)(3) – (y – 2)(1) + (z – 3)(1) = 0 3x – 3 – y + 2 + z – 3 = 0 3x – y + z – 4 = 0, is the required plane equation Now, given O(0,0,0) & P(2,–1,1) Plane is 3x – y + z – 4 = 0 Let O’ & P’ be the foot of perpendiculars.
Q. 10: The contra positive of the statement “If you will work, you will earn money” is:
(A) If you will not earn money, you will not work
(B) You will earn money, if you will not work
(C) If you will earn money, you will work
(D) To earn money, you need to work
Answer: (A) Contrapositive of p → q is ~q →~p p: you will work q: you will earn money ~q: you will not earn money ~p: you will not work ~q →~p: if you will not earn money, you will not work
Q. 11: The shortest distance between the line x – y = 1 and the curve x2 = 2y is:
(A) 1/2
(B) 0
(C) 1/(2√2)
(D) 1/√2
Answer: (C) Shortest distance must be along common normal m1 (slope of line x–y = 1) = 1 ⇒slope of perpendicular line =–1 m2 = 2x/2 = x ⇒ m2 = h ⇒slope of normal = –1/h –1/h=–1 ⇒h=1 So, the point is (1,1/2) Therefore, Shortest distance (D) =|(1–1/2–1)/√(1+1)|=1/(2√2)
Q. 12: Let A be a set of all 4-digit natural numbers whose exactly one digit is 7. Then the probability that a randomly chosen element of A leaves remainder 2 when divided by 5 is:
(A) 1/5
(B) 2/9
(C) 97/297
(D) 122/297
Answer: (C) Total cases = 4C1 × 9 × 9 × 9 – 3C1 × 9 × 9 (as 4 digit number having 0 at thousands place have to be excluded) For a number to have a remainder 2 when divided by 5 it’s unit digit should be 2 or 7 Case 1: when unit digit is 2 Number of four-digit number = 3C1×9×9 – 2C1×9 Case 2: when unit digit is 7 Number of four digits number = 8 × 9 × 9 So total number favorable cases = 3×92 – 2×9 + 8×92 Required Probability = ((3×9×9)–(2×9)+(8×9×9))/((4×93) – (3×92)) = 97/297
Q. 13: cosec[2 cot–1( 5) + cos–1(4/5) ] is equal to:
(A) 75/56
(B) 65/56
(C) 56/33
(D) 65/33
Answer: (B) cosec[2 cot–1( 5) + cos–1(4/5) ]
Q. 14: If 0 < x, y < π and cos x + cos y – cos (x + y) = 3/2, then sin x + cos y is equal to:
(A) (1+√3)/2
(B) (1–√3)/2
(C) √3/2
(D) 1/2
Answer: (A)
Q. 15: If α, β∈ R are such that 1 – 2i (here i2 = –1) is a root of z2 + αz + β = 0, then (α – β) is equal to:
(A) 7
(B) -3
(C) 3
(D) -7
Answer: (D) 1-2i is a root of z2 + αz + β = 0. (1-2i) 2 + α(1-2i)+β=0 ⇒1-4-4i+α-2iα+β=0 ⇒(α+β-3)-i(4+2α)=0 ⇒α+β-3=0 & 4+2α=0 So, α=-2, β=5 Therefore, α-β=-7
Q. 16: If , then
(A) 1/(I2 + I4 ) , 1/(I3 + I5 ), 1/(I4 + I6) are in G.P.
(B) 1/(I2 + I4 ) , 1/(I3 + I5 ), 1/(I4 + I6) are in A.P.
(C) I2 + I4 , I3 + I5 , I4 + I6) are in A.P.
(D) I2 + I4 , I3 + I5 , I4 + I6) are in G.P.
Answer: (B)
In+2 + In = 1/(n+1) I2 + I4 =1/3, I3 + I5 =1/4 and I4 + I6 =1/5 So, 1/(I2 + I4 ) , 1/(I3 + I5 ) and 1/(I4 + I6) are in A.P.
Q. 17: If for the matrix, A= AAT = I2, then the value of α4 + β4 is:
(A) 1
(B) 3
(C) 2
(D) 4
Answer: (A)
Q. 18: Let x denote the total number of one-one functions from a set A with 3 elements to a set B with 5 elements and y denote the total number of one-one functions from the set A to the set A × B. Then:
(A) y = 273x
(B) 2y = 91x
(C) y = 91x
(D) 2y = 273x
Answer: (B) Number of elements in A = 3 Number of elements in B = 5 Number of elements in A × B = 15 Number of one-one function is x = 5 × 4 × 3 ⇒x = 60 Number of one-one function is y = 15 × 14 × 13 ⇒y = 15 × 4 × 14/4 × 13 ⇒y = 60 × 7/2 × 13 ⇒2y = (13)(7x) ⇒2y = 91x
Q. 19: Let α and β be the roots of x2 – 6x – 2 = 0. If an = αn – βn for n ≥ 1, then the value of (a10–2a8)/(3a9 ) is:
(A) 4
(B) 1
(C) 2
(D) 3
Answer: (C) α and β be the roots of x2–6x–2=0. So, α2–6α–2=0⇒α2–2=6α β2–6β–2=0⇒β2–2=6β Now,
Q. 20: Let A be a 3 × 3 matrix with det(A) = 4. Let Ri denote the ith row of A. If a matrix B is obtained by performing the operation R2→ 2R2 + 5R3 on 2A, then det(B) is equal to:
(A) 64
(B) 16
(C) 80
(D) 128
Answer: (A)
Section B
Q. 1: Let . If the area of the parallelogram whose adjacent sides are represented by the vectors a￫and a￫ is 8√3 square units, then
a￫.b￫is equal to ______.
Answer: 2 Area of parallelogram =
|a￫ × b￫|
(64)(3) = 16α2 + 64 + 16α2 ⇒ α2 = 4 Now,
a￫.b￫ = 3 – α2 + 3 = 6 – α2 = 6 – 4 = 2
Q. 2: If the curve y = y(x) represented by the solution of the differential equation (2xy2 – y)dx + xdy = 0, passes through the intersection of the lines, 2x – 3y = 1 and 3x + 2y = 8, then |y(1)| is equal to ______.
Answer:1 Given, (2xy2 – y)dx + xdy = 0 ⇒ dy/dx + 2y2 – y/x = 0 ⇒ (–1/y2) . dy/dx + (1/y) (1/x) = 2 Let, 1/y = z (–1/y2) dy/dx = dz/dx ⇒dz/dx + z(1/x) = 2 As it passes through P(2, 1) [Point of intersection of 2x – 3y = 1 and 3x + 2y = 8] Therefore, 2/1 = 4 + c ⇒c = –2 ⇒x/y = x2 – 2 Put x = 1 1/y = 1 – 2 = –1 ⇒y(1) = –1 ⇒|y(1)| = 1
Q. 3: If limx→0ax−(e4x−1) / ax(e4x−1) exists and is equal to b, then the value of a – 2b is ______.
Answer: 5
limx→0ax−(e4x−1) / ax(e4x−1)
Applying L’Hospital Rule
limx→0a−4e4x / a(e4x−1)+ax(4e4x) So for limit to exist,a=4 Applying L’Hospital Rule
Q. 4: A line L passing through origin is perpendicular to the lines
If the co-ordinates of the point in the first octant on L2 at the distance of √17 from the point of intersection of L and L1 are (a, b, c), then 18(a+b+c) is equal to ______.
Answer: 44 D.R. of L is parallel to (L1 × L2) ⇒ (-2, 3, -2) Equation of l : x/2 = y/(–3) = z/2 Solving L & L1 (2λ, –3λ, 2λ) = (µ + 3, 2µ – 1, 2µ + 4) µ = – 1 , λ = 1 So, intersection point P(2, –3, 2) Let, Q(2ν + 3, 2ν + 3, ν + 2) be required point on L2
Q. 5: A function f is defined on [–3,3] as
where [x] denotes the greatest integer ≤ x. The number of points, where f is not differentiable in (–3,3) is ______.
Answer: (5) Points of non-differentiable in (–3, 3) are at x = –2, –1, 0, 1, 2. i.e. 5 points.
Q. 6: A line is a common tangent to the circle (x – 3)2 + y2 = 9 and the parabola y2 = 4x. If the two points of contact (a, b) and (c, d) are distinct and lie in the first quadrant, then 2(a+c) is equal to ______.
Answer: 9 Sol. Circle: (x – 3)2+ y2= 9 Parabola: y2 = 4x Let common tangent equation be y = mx + a/m ⇒y = mx + 1/m ⇒m2x – my + 1 = 0 The above line is also tangent to circle (x – 3)2 + y2= 9 Therefore, the perpendicular from (3, 0) to line = 3 ⇒|(3m2 – 0 + 1)/√(m2+m4 )| = 3 ⇒(3m2 + 1)2 = 9(m2 + m4) ⇒m = ±1/√3 Tangent is y = 1/√3x + √3 (it will be used) => m = 1/√3 or y = –1/√3x – √3 (rejected) For Parabola, point of contact is (a/m2 , 2a/m)= (3, 2√3) = (c, d) Solving Circle (x – 3)2 + y2 = 9 & line equation y = (1/√3) x + √3 (x – 3)2 + ((1/√3) x + √3)2 = 9 ⇒x2 + 9 – 6x + (1/3)x2 + 3 + 2x = 9 ⇒(4/3)x2 – 4x + 3 = 0 ⇒x = 3/2=a ∴ 2(a + c) = 2(3/2+3)= 9
Q. 7: The value of is ______.
Answer: 19
x2-x-2=(x–2)(x+1)
Q. 8: If the remainder when x is divided by 4 is 3, then the remainder when (2020+x)2022 is divided by 8 is ______.
Answer: 1 Let x = 4k + 3 (2020 + x)2022 = (2020 + 4k + 3)2022 = (2024 + 4k – 1)2022 = (4A – 1)2022 =2022C0(4A)2022(–1)0 + 2022C1(4A)2021(–1)1 + ……+2022C2021(4A)1(–1)2021+ 2022C2022(4A)0(–1)2022 Which will be of the form 8λ+1 So, Remainder is 1.
Q. 9: If the curves x = y4 and xy = k cut at right angles, then (4k)6 is equal to ______.
Answer: 4
Q. 10: The total number of two digit numbers ‘n’, such that 3n+7n is a multiple of 10, is ______.
Answer: 45 Let n = 2t; t ∈ N 3n = 32t = (10 – 1)t =10p + (–1)t = 10p ± 1 If n = even then 7n + 3n will not be multiple of 10 So, if n is odd then only 7n + 3n will be a multiple of 10 Therefore, n = 11,13,15,………..,99 Therefore, the answer is 45
Choose the correct answer from the options given below:
(A) (A) → (ii), (B) →(iii), (C) → (iv), (D) → (i)
(B) (A) →(i), (B) → (ii), (C) →(iv), (D) → (iii)
(C) (A) → (iii), (B) → (ii), (C) →(iv), (D) →(i)
(D) (A) → (iii), (B) → (iv), (C) →(ii), (D) →(i)
Answer: (A) K.E. = [ML2T–2] P (Linear momentum) = [MLT–1] h (Planck’s constant) = [ML2T–1] V (Electric potential) = [ML2T–3I–1]
Q. 2: The pitch of the screw gauge is 1 mm and there are 100 divisions on the circular scale. When nothing is put in between the jaws, the zero of the circular scale lines 8 divisions below the reference line. When a wire is placed between the jaws, the first linear scale division is clearly visible while the 72nd division on circular scale coincides with the reference line. The radius of the wire is:
(A) 1.64 mm
(B) 1.80 mm
(C) 0.82 mm
(D) 0.90 mm
Answer: (C) Least count. = pitch/no. of div. = 1 mm/100 = 0.01 mm +ve zero error = 8 × L.C. = +0.08 mm Measured reading = 1mm + 72 × L.C. = 1mm + 0.72 mm = 1.72 mm True reading = 1.72 – 0.08 = 1.64 mm Radius = 1.64/2 = 0.82 mm
Q. 3: If the time period of a two meter long simple pendulum is 2 s, the acceleration due to gravity at the place where pendulum is executing S.H.M. is:
(A) 2π2 ms–2
(B) 16 m/s2
(C) 9.8 ms–2
(D) π2 ms–2
Answer: (A) T = 2π√(l/g) T2 = 4π2l/g g = 4π2l/T2 = 4π2×2/(2)2 = 2π2 ms–2
Q. 4: An α particle and a proton are accelerated from rest by a potential difference of 200 V. After this, their de Broglie wavelengths are λα and λp respectively. The ratio λp/λα is:
Q. 5: Given below are two statements: one is labelled as Assertion A and the other is labelled as reason R.
Assertion A: The escape velocities of planet A and B are same. But A and B are of unequal masses.
Reason R: The product of their masses and radii must be same. M1R1 = M2R2
In the light of the above statements, choose the most appropriate answer from the options given below:
(A) Both A and R are correct but R is NOT the correct explanation of A
(B) A is correct but R is not correct
(C) Both A and R are correct and R is the correct explanation of A
(D) A is not correct but R is correct
Answer: (B) Ve = escape velocity ve = √(2GM/R) So, for same ve, M1/R1 = M2/R2 A is true but R is false
Q. 6: A diatomic gas, having Cp = (7/2)R and Cv = (5/2)R, is heated at constant pressure. The ratio dU : dQ : dW
(A) 3 : 7 : 2
(B) 5 : 7 : 2
(C) 5 : 7 : 3
(D) 3 : 5 : 2
Answer: (B) Since the gas is diatomic in nature and the process is isobaric, we have Cp = (7/2)R Cv = (5/2)R dU = nCvdT dQ = nCpdT dW = nRdT dU : dQ : dW Cv : Cp : R (5/2) R : (7/2) R : R 5 : 7 : 2
Q. 7: Statement I: A speech signal of 2 kHz is used to modulate a carrier signal of 1 MHz. The bandwidth requirement for the signal is 4 kHz.
Statement II: The side band frequencies are 1002 kHz and 998 kHz.
In the light of the above statements, choose the correct answer from the options given below:
(A) Both statement I and statement II are false
(B) Statement I is false but statement II is true
(C) Statement I is true but statement II is false
(D) Both statement I and statement II are true
Answer: (D) Side band = (fc – fm) to (fc + fm) = (1000 – 2) kHz to (1000 + 2) kHz = 998 kHz to 1002 kHz Band width = 2fm = 2 ×2kHz = 4 kHz Both statements are true.
Q. 8: The current (i) at time t = 0 and t = ∞ respectively for the given circuit is:
(A) 18E/55,5E/18
(B) 5E/18,18E/55
(C) 5E/18,10E/33
(D) 10E/33,5E/18
Answer: (C) At t = 0, inductor is open Initial-state equivalent of the circuit shown in figure -1 is Req = 6×9/(6+9) = 54/15 I (t = 0) = E×15/54 = 5E/18 At t = ∞, For steady state inductor is replaced by plane wire Steady state equivalent of the circuit shown in figure-1 is Equivalent circuit diagram is given by Req =1×4/(1+4) + 5×5/(5+5) = 4/5 + 5/2 = (8 + 25)/10 = 33/10 I = E/Req = 10E/33
Q. 9: Two satellites A and B of masses 200 kg and 400 kg are revolving round the earth at height of 600 km and 1600 km respectively.
If TA and TB are the time periods of A and B respectively then the value of TB – TA:
[Given: Radius of earth = 6400 km, mass of earth = 6×1024 kg]
(A) 4.24× 102 s
(B) 3.33 × 102 s
(C) 1.33 × 103 s
(D) 4.24 × 103 s
Answer: (C) V = √(GMe/r)
T = 2𝜋r/ (√(GMe/r)) = 2𝜋r/ (√(r/GMe))
T =√(4𝜋2r3/GMe) = √(4𝜋2r3/GMe) T2 – T1 = √(4𝜋2(8000×103)3/G×6×1024 ) – √(4𝜋2(7000×103)3/G×6×1024 ) ≅ 1.33 ×103 s
Q. 10: An engine of a train, moving with uniform acceleration, passes the signal post with velocity u and the last compartment with velocity v. The velocity with which middle point of the train passes the signal post is:
(A) √(v2 – u2)/2
(B) (v – u)/2
(C) √(v2 + u2)/2
(D) (u + v)/2
Answer: (C) a = uniform acceleration u = velocity of first compartment v = velocity of last compartment l = length of train v2 = u2 + 2as (3rd equation of motion) v2 = u2 + 2al …..(1) v2middle = u2 + 2al/2 ∴ v 2middle = u2 + al ….(2) From equation (1) and (2) v2middle = u2 + (v2 – u2)/2 = (v2 + u2)/2 ∴ vmiddle = √(v2 + u2)/2
Q. 11: A 5V battery is connected across the points X and Y. Assume D1 and D2 to be normal silicon diodes. Find the current supplied by the battery if the +ve terminal of the battery is connected to point X.
(A) ~0.86 A
(B) ~0.5 A
(C) ~0.43 A
(D) ~1.5 A
Answer: (C) Since silicon diode is used so 0.7 Volt is drop across it, only D1 will conduct so current through cell. I = (5 – 0.7)/10 = 0.43 A
Q. 12: A solid sphere of radius R gravitationally attracts a particle placed at 3R from its centre with a force F1. Now a spherical cavity of radius R/2 is made in the sphere (as shown in figure) and the force becomes F2. The value of F1: F2 is:
Q. 13: A student is performing the experiment of resonance column. The diameter of the column tube is 6 cm. The frequency of the tuning fork is 504 Hz. Speed of the sound at the given temperature is 336 m/s. The zero of the metre scale coincides with the top end of the resonance column tube. The reading of the water level in the column when the first resonance occurs is:
(A) 13 cm
(B) 14.8 cm
(C) 16.6 cm
(D) 18.4 cm
Answer: (B) λ = v/f = 336/504 = 66.66cm λ/4 = l + e = l + 0.3d = l + 1.8 16.66 = l + 1.8 cm l = 14.86 cm
Q. 14: A proton, a deuteron and an α particle are moving with the same momentum in a uniform magnetic fiel(D) The ratio of magnetic forces acting on them is _______ and their speeds are in the ratio______.
(A) 2 : 1 : 1 and 4 : 2 : 1
(B) 1 : 2 : 4 and 2 : 1 :1
(C) 1 : 2 : 4 and 1 : 1 : 2
(D) 4 : 2 : 1 and 2 : 1 : 1
Answer: (A) As v = p/m & F = qvB ∴ F = qpB/m F1 = qpB/m, v1 = p/m F2 = qpB/2m, v2 = p/2m F3 = 2qpB/4m, v3 = p/4m F1 : F2 : F3 & V1 : V2 : V3 1 : ½ : ½ & 1 : ½ :¼ 2 : 1 : 1 & 4 : 2 : 1
Q. 15: Given below are two statements: one is labelled as Assertion A and the other is labelled as Reason R.
Assertion A: When a rod lying freely is heated, no thermal stress is developed in it.
Reason R: On heating, the length of the rod increases
In the light of the above statements, choose the correct answer from the options given below:
(A) A is true but R is false
(B) Both A and R are true and R is the correct explanation of A
(C) Both A and R are true but R is NOT the correct explanation of A
(D) A is false but R is true
Answer: (C) If a rod is free and it is heated then there is no thermal stress produced in it. The rod will expand due to increase in temperature. So, both A & R are true.
Q. 16: In an octagon ABCDEFGH of equal side, what is the sum of
Answer: (A)
Q. 17: Two radioactive substances X and Y originally have N1 and N2 nuclei respectively. Half-life of X is half of the half-life of Y. After three half-lives of Y, numbers of nuclei of both are equal. The ratio N1/N2 will be equal to:
(A) 8/1
(B) 1/8
(C) 3/1
(D) 1/3
Answer: (A) After n half-life no of nuclei undecayed = No/2n Given, t(1/2)x= t(1/2y)/2 So 3 half-life of y = 6 half-life of x Given, Nx = Ny after 3t(½)y N1/26 = N2/23 N1/N2 = 26/23 = 23= 8/1
Q. 18: The angular frequency of alternating current in a L-C-R circuit is 100 rad/s. The components connected are shown in the figure. Find the value of inductance of the coil and capacity of condenser.
(A) 0.8 H and 250 μF
(B) 0.8 H and 150 μF
(C) 1.33 H and 250 μF
(D) 1.33 H and 150 μF
Answer: (A) Since key is open, circuit is series L-C-R circuit 15 = irms (60) ∴irms = ¼ A Now, 20 = ¼ XL = ¼ (ωL) ∴ L = 4/5 = 0.8 H & 10 = ¼ 1/(100C) C = (1/4000) F = 250 μF
Q. 19: Two coherent light sources having intensities in the ratio 2x produce an interference pattern. The ratio (Imax – Imin)/(Imax + Imin) will be:
Q. 20: Magnetic fields at two points on the axis of a circular coil at a distance of 0.05 m and 0.2 m from the centre are in the ratio 8 : 1. The radius of coil is ______
Q. 1: The potential energy (U) of a diatomic molecule is a function dependent on r (interatomic distance) as U= α/r10– β/r5 – 3. Where, a and b are positive constants. The equilibrium distance between two atoms will (2α/β)a/(B) Where a =_______
Answer: 1 F = -dU/dr F = -[-10α/r11 + 5β/r6] for equilibrium, F = 0 10α/r11 = 5β/r6 2α/β = r5 r = (2α/β)1/5 a = 1
Q. 2: A small bob tied at one end of a thin string of length 1m is describing a vertical circle so that the maximum and minimum tension in the string are in the ratio 5 : 1. The velocity of the bob at the highest position is ______ m/s. (take g = 10 m/s2)
Q. 3:In a certain thermodynamic process, the pressure of a gas depends on its volume as kV3. The work done when the temperature changes from 1000 C to 3000 C will be ____ nR, where n denotes the number of moles of a gas.
Answer: 50 P = kv3 pv–3 = k x = –3 w = nR(T1 – T2)/(x – 1) = nR(100-300)/(-3 -1) = nR(-200)/-4 = 50nR
Q. 4: In the given circuit of potentiometer, the potential difference E across AB (10 m length) is larger than E1 and E2 as well. For key K1 (closed), the jockey is adjusted to touch the wire at point J1 so that there is no deflection in the galvanometer. Now the first battery (E1) is replaced by the second battery (E2) for working by making K1 open and E2 close(D) The galvanometer gives then null deflection at J2. The value of E1/E2 is the smallest fraction of a/b, Then the value of a is ____.
Answer: 1 E1/E2 = l1/l2 = 3×100 cm + (100 – 20)cm)/(7×100 cm + 60 cm) = 380/760 = ½ = a/b a = 1
Q. 5: The same size images are formed by a convex lens when the object is placed at 20 cm or at 10 cm from the lens. The focal length of a convex lens is ______ cm.
Answer: 15 1/v – 1/u = 1/f …(1) m = v/u ….(2) from (1) and (2) we get m = f/(f + u) Given conditions m1 = -m2 f/(f – 10) = -f/(f – 20) f – 20 = –f + 10 2f = 30 f = 15 cm
Q. 6: A transmitting station releases waves of wavelength 960 m. A capacitor of 256 μF is used in the resonant circuit. The self inductance of coil necessary for resonance is _____ × 10–8H.
Answer: 10 At resonance ωr = 1/√(Lc) ∴ 2πf = 1/√(Lc) ∴ 4π2c2/λ2 = 1/Lc ∴4π2×3×108×3×108/960×960 = 1/L×2.56×10-6 L = 375×960/10-6×4×π2×9×1016 = 103/1010 = 10–7 H = 10 × 10–8 H
Q. 7: The electric field in a region is given by . The ratio of flux of reported field through the rectangular surface of area 0.2 m2 (parallel to y-z plane) to that of the surface of area 0.3 m2(parallel to x-z plane) is a : 2, where a = ________
[Here i, j and k are unit vectors along x, y and z-axes respectively]
Answer: 1
Therefore, a = 1
Q. 8: 512 identical drops of mercury are charged to a potential of 2 V each. The drops are joined to form a single drop. The potential of this drop is ____ V.
Answer: 128 Let charge on each drop = q Radius = r V = kq/r 2 = kq/r Radius of bigger 4πR3/3 = 512 × (4/3) πr3 R = 8r V = k(512)q/R = (512/8) (kq/r) = (512/8) × 2 = 128 V
Q. 9: A monatomic gas of mass 4.0 u is kept in an insulated container. Container is moving with a velocity 30 m/s. If the container is suddenly stopped then the change in temperature of the gas (R=gas constant) is x/3R. Value of x is ______.
Q. 10: A coil of inductance 2 H having negligible resistance is connected to a source of supply whose voltage is given by V = 3t volt. (where t is in second). If the voltage is applied when t = 0, then the energy stored in the coil after 4 s is _______ J.
Answer: 144 L di/dt = ε = 3t L ∫di = 3 ∫t dt Li = 3t2/2 i = 3t2/2L Energy stored in the coil, E = ½ Li2 = ½ L (3t2/2L)2 = ½ × 9t4/4L = 9/8 × (4)4/2 = 144 J
Chemistry
SECTION A
Q. 1. The hybridization and magnetic nature of [Mn(CN)6]4– and [Fe(CN)6]3–, respectively are:
(A) d2sp3 and paramagnetic
(B) sp3d2 and paramagnetic
(C) d2sp3 and diamagnetic
(D) sp3d2 and diamagnetic
Answer: (A)
Q. 2. Identify A and B in the chemical reaction.
Answer: (D)
Q. 3. Compound(s) which will liberate carbon dioxide with sodium bicarbonate solution is/are:
(A) 2 and 3 only
(B) 1 only
(C) 2 only
(D) 3 only
Answer: (A) Compounds that are more acidic than H2CO3, gives CO2 gas in reaction with NaHCO3. Compound B i.e. Benzoic acid and compound C i.e. picric acid both are more acidic than H2CO3.
Q. 4. Ellingham diagram is a graphical representation of:
(A) ΔG vs T
(B) (ΔG – TΔS) vs T
(C) ΔH vs T
(D) ΔG vs P
Answer: (A) Ellingham diagram tells us about the spontaneity of a reaction with temperature.
Q. 5. Which of the following equations depicts the oxidizing nature of H2O2?
(A) Cl2 + H2O2 → 2HCl + O2
(B) KlO4 + H2O2 → KlO3 + H2O + O2
(C) 2l– + H2O2 + 2H+ → I2 + 2H2O
(D) I2 + H2O2 + 2OH– → 2I– + 2H2O + O2
Answer: (C) 2l– + H2O2 + 2H+ → I2 + 2H2O Oxygen reduces from –1 to –2 So, its reduction will take place. Hence it will behave as oxidising agent or it shows oxidising nature.
Q. 6. In Freundlich adsorption isotherm at moderate pressure, the extent of adsorption (x/m) is directly proportional to Px. The value of x is:
(A) ∞
(B) 1
(C) zero
(D) 1/n
Answer: (D) x / m = px The formula is x / m = px Hence, x= 1 / n. The value of ‘n’ is any natural number.
Q. 7. According to molecular orbital theory, the species among the following that does not exist is:
(A) He2–
(B) He2+
(C) O22-
(D) Be2
Answer: (D) B.O. of Be2 is zero, so it does not exist.
Q. 8. In which of the following pairs, the outermost electronic configuration will be the same?
(A) Fe2+ and Co+
(B) Cr+ and Mn2+
(C) Ni2+ and Cu+
(D) V2+ and Cr+
Answer: (B) Cr+ → [Ar]3d5 Mn2+ → [Ar]3d5
Q. 9. Given below are two statements:
Statement-I: An allotrope of oxygen is an important intermediate in the formation of reducing smog.
Statement-II: Gases such as oxides of nitrogen and Sulphur present in the troposphere contribute to the formation of photochemical smog.
(A) Statement I and Statement II are true
(B) Statement I is true about Statement II is false
(C) Both Statement I and Statement II are false
(D) Statement I is false but Statement II is true
Answer: (C) Reducing smog acts as a reducing agent and the reducing character is due to presence of sulphur dioxide and carbon particles.
Q. 10. The plots of radial distribution functions for various orbitals of hydrogen atom against ‘r’ are given below:
(A) (4)
(B) (2)
(C) (1)
(D) (3)
Answer: (A) 3s orbital Number of radial nodes = n – λ – 1 For 3s orbital n = 3, λ = 0 Number of radial nodes = 3 – 0 – 1 = 2 It is correctly represented in graph of option 4
Q. 11. Identify A in the given chemical reaction.
Answer: (D)
Aromatization reaction or hydroforming reaction.
Q. 12. Given below are two statements:
Statement-I: CeO2 can be used for oxidation of aldehydes and ketones.
Statement-II: Aqueous solution of EuSO4 is a strong reducing agent.
(A) Statement I is true, statement II is false
(B) Statement I is false, statement II is true
(C) Both Statement I and Statement II are false
(D) Both Statement I and Statement II are true
Answer: (D) CeO2 can be used as an oxidizing agent like seO2. Similarly, EuSO4 is used as a reducing agent.
Q. 13. The major product of the following chemical reaction is:
(A) (CH3CH2CO)2O
(B) CH3CH2CHO
(C) CH3CH2CH3
(D) CH3CH2CH2OH
Answer: (B)
Q. 14. Complete combustion of 1.80 g of an oxygen-containing compound (CxHyOz) gave 2.64 g of CO2 and 1.08 g of H2O. The percentage of oxygen in the organic compound is:
(C) Terminal B–H bonds have less p-character when compared to bridging bonds.
(D) The two B–H–B bonds are not of the same length.
Answer: (C) The terminal bond angle is greater than that of bridge bond angle Bond angle ∝ S-character
α 1/ p−character
Q. 16. Which of the glycosidic linkage galactose and glucose is present in lactose?
(A) C-1 of glucose and C-6 of galactose
(B) C-1 of galactose and C-4 of glucose
(C) C-1 of glucose and C-4 of galactose
(D) C-1 of galactose and C-6 of glucose
Answer: (B)
Q. 17. Which one of the following reactions will not form acetaldehyde?
Answer: (A)
Q. 18. Which of the following reactions will not give p-amino azobenzene?
(A) 2 only
(B) 1 and 2
(C) 3 only
(D) 1 only
Answer: (A)
Q. 19. The solubility of AgCN in a buffer solution of pH = 3 is x. The value of x is: [Assume: No cyano complex is formed; Ksp(AgCN) = 2.2 × 10–16 and Ka (HCN) = 6.2 × 10–10]
(A) 0.625 × 10-6
(B) 1.6 × 10-6
(C) 2.2 × 10-16
(D) 1.9 × 10-5
Answer: (D) Let solubility is x AgCN ⇌ Ag+ + CN–Ksp = 2.2 × 10–16 x x H+ + CN– ⇌ HCN
(A) Buna-S is a synthetic and linear thermosetting polymer
(B) Neoprene is an addition copolymer used in plastic bucket manufacturing
(C) Synthesis of Buna-S needs nascent oxygen
(D) Buna-N is a natural polymer
Answer: (C) Synthesis of Buna-S needs nascent oxygen.
SECTION B
Q. 1. A car tire is filled with nitrogen gas at 35 psi at 27°C. It will burst if pressure exceeds 40 psi. The temperature in °C at which the car tyre will burst is ______. (Rounded-off to the nearest integer)
Q. 2. The reaction of cyanamide, NH2CN(s) with oxygen was run in a bomb calorimeter and ΔU was found to be –742.24 kJ mol–1. The magnitude of ΔH298 for the reaction NH2CN (s) + 3/2 O2 (g) → N2 (g) + O2 (g) + H2O(l) is______kJ. (Rounded off to the nearest integer). [Assume ideal gases and R = 8.314 J mol–1 K–1]
Q. 3. The ionization enthalpy of Na+ formation from Na(g) is 495.8 kJ mol–1, while the electron gain enthalpy of Br is –325.0 kJ mol–1. Given the lattice enthalpy of NaBr is –728.4 kJ mol–1. The energy for the formation of NaBr ionic solid is (–)_____ × 10–1 kJ mol–1.
Answer: 7 All carbon atoms in benzaldehyde are sp2 hybridized.
Q. 5. Among the following, the number of halide(s) which is/are inert to hydrolysis is ______.
(A) BF3
(B) SiCl4
(C) PCl5
(D) SF6
Answer: 1 Due to crowding, SF6 is not hydrolyzed.
Q. 6. In basic medium CrO42– oxidizes S2O32– to form SO24 and itself changes into Cr(OH)4–. The volume of 0.154 M CrO42– required to react with 40 mL of 0.25 M S2O32– is ______ mL. (Rounded-off to the nearest integer)
Answer: 173 mL 17H2O + 8CrO4 + 3S2O3 → 6SO4 + 8Cr(OH)4– + 2OH– Applying mole-mole analysis 0.154 × v / 8 = 40 × 0.25 / 3 V = 173 mL
Q. 7. 1 molal aqueous solution of an electrolyte A2B3 is 60% ionise(D) The boiling point of the solution at 1 atm is _____ K. (Rounded-off to the nearest integer). [Given Kb for (H2O) = 0.52 K kg mol–1]
Q. 8. Using the provided information in the following paper chromatogram:
The calculated Rf value of A ______ × 10–1.
Answer: 4
Rr = Distance Travelled By Solvent / Distance Travelled By Compound
On chromatogram distance travelled by compound is → 2 cm Distance travelled by solvent = 5 cm So Rf = 2 / 5 = 4 × 10–1 = 0.4
Q. 9. For the reaction, aA + bB → cC + dD, the plot of log k vs 1/T is given below:
The temperature at which the rate constant of the reaction is 10–4s–1 is ________ K. [Rounded off to the nearest integer)
[Given: The rate constant of the reaction is 10–5 s–1 at 500 K]
Answer: 526 K
log10 K = log10 A – E0/2.303RT
Slope = E0/2.203 = -10000
log10 K1/K2 = E0/2.203 × (1/T1 – 1/T2)
log10 10-4/10-5 = 1000 × (1/500 -1/T)
1 = 1000 × (1/500 -1/T)
1/10000 = 1/500 = 1/T
1/T = 1/500 = 1/10000
1/T = 20-1 / 10000 = 19/10000
T = 10000 / 19 T= 526 K
Q. 10. A 0.4g mixture of NaOH, Na2CO3 and some inert impurities was first titrated with N / 10HCl using phenolphthalein as an indicator, 17.5 mL of HCl was required at the end point. After this methyl orange was added and titrate(D) 1.5 mL of the same HCl was required for the next end point. The weight percentage of Na2CO3 in the mixture is ______. (Rounded-off to the nearest integer)
Q. 1: A man is observing, from the top of a tower, a boat speeding towards the tower from a certain point A, with uniform spee(D) At that point, the angle of depression of the boat with the man’s eye is 30° (Ignore man’s height). After sailing for 20 seconds towards the base of the tower (which is at the level of water), the boat has reached point B, where the angle of depression is 45°. Then the time taken (in seconds) by the boat from B to reach the base of the tower is :
(A) 10(√3-1)
(B) 10√3
(C) 10
(D) 10(√3+1)
Answer: (D) x + y = √3h ….…….(1) Also, h/y = tan 45o h = y …….(2) put in (1) x + y = √3y x = (√3 – 1)y x/20=v (speed) Therefore, time taken to reach Foot from B = y/V = x/(√3-1)x . 20 = 10 (√3 + 1)
Q. 2:
(A) xyz = 4
(B) xy – z = (x + y)z
(C) xy + yz + zx = z
(D) xy + z = (x+y)z
Answer: (D)
Q. 3: The equation of the line through the point (0,1,2) and perpendicular to the line (x-1)/2 = (y+1)/3 = (z-1)/-2 is :
(A) x/-3 = (y-1)/ 4 = (z-2)/3
(B) x/3 = (y-1)/ 4 = (z-2)/3
(C) x/3 = (y-1)/ -4 = (z-2)/3
(D) x/3 = (y-1)/ 4 = (z-2)/-3
Answer: (A)
(x-1)/2 = (y+1)/3 = (z-1)/-2 =λ Any point on this line (2λ + 1, 3λ – 1, -2λ + 1) Direction ratio of given line d1 ≡ (2, 3, -2) Direction ratio of the line to be found d2 ≡(2λ+1,3λ−2,−2λ−1)
∴ d1 d2 = 0
λ = 2 / 17 Direction ratio of line (21, -28, -21) ≡ (3, -4, -3) ≡ (-3, 4, 3)
Q. 4: The coefficients a,b and c of the quadratic equation, ax2 + bx + c = 0 are obtained by throwing a dice three times. The probability that this equation has equal roots is :
(A) 1/54
(B) 1/72
(C) 1/36
(D) 5/216
Answer: (D) ax2 + bx + c = 0 a,b,c ∈ {1,2,3,4,5,6} n(s) = 6 × 6 × 6 = 216 D=0 ⇒ b2 = 4ac ac = b2/4, If b = 2, ac = 1 ⇒ a = 1,c = 1 If b = 4, ac = 4 ⇒ a = 1, c = 4 a = 4, c = 1 a = 2, c = 2 If b = 6, ac = 9 ⇒ a = 3, c = 3 Therefore, probability = 5/216
Q. 5: Let α be the angle between the lines whose direction cosines satisfy the equations l +m – n = 0 and l2 + m2 – n2 = 0. Then the value of sin4 α + cos4 α is :
(A) 3/4
(B) 1/2
(C) 5/8
(D) 3/8
Answer: (C) Given that l+m=n ….(1) l2 + m2 – n2 = 0 ….(2) Squaring equation (1) l2 + m2 + 2lm = n2 ….(3) From equations (2) and (3) lm=0 ⇒ l = 0 or m = 0 Case (1) : l = 0 ⇒ m=n ⇒ l2 + m2 + n2 = 0 ⇒ m=n = ±1/(√2 ) ∴(l,m,n) = (0, 1/√2, 1/√2) or (0, -1/√2, -1/√2) Case (2): m = 0 ⇒ l = n ⇒ l2 + m2 + n2 = 0
Answer: (C) Using trig identities, sin2A = 2sinAcosA and 1-cos2A = 2sin2A
Q. 7: The statement A→ (B → A) is equivalent to:
(A) A → (A ᐱ B)
(B) A → (A ᐯ B)
(C) A → (A → B)
(D) A → (A ↔ B)
Answer: (B) A → (B → A) ⇒ A → (∼ B ᐯ A) ⇒ ∼ A ᐯ (∼ B ᐯ A) ⇒ ∼ B ᐯ (∼ A ᐯ A) ⇒ ∼ B ᐯ t = t (tautology) From options: (B) A → (A ᐯ B) ⇒ ∼ A ᐯ (A ᐯ B) ⇒ (∼ A ᐯ A) ᐯ B ⇒ t ᐯ B ⇒ t
Q. 8: The integer k, for which the inequality x2 – 2(3k – 1)x + 8k2 – 7 >0 is valid for every x in R is :
(A) 3
(B) 2
(C) 4
(D) 0
Answer: (A) D < 0 (2(3k-1))2 – 4(8k2 – 7) < 0 4(9k2 – 6k + 1) – 4(8k2 – 7) < 0 k2 – 6 k + 8 < 0 (k–4)(k–2)<0 2< k < 4 then k = 3
Q. 9: All possible values of θ ∈ [0, 2π] for which sin2θ + tan2θ > 0 lie in
a.(0,π/2)∪(π,3π/2)
(B) (0,π/4)∪(π/2,3π/4)∪(π,5π/4)∪(3π/2,7π/4)
(C) (0,π/2)∪(π/2,3π/4)∪(π,7π/4)
(D) (0,π/4)∪(π/2,3π/4)∪(3π/2,11π/6)
Answer: (B)
Q. 10: The image of the point (3,5) in the line x – y + 1 = 0, lies on :
(A) (x – 2)2 + (y – 4) 2 =4
(B) (x – 4) 2 + (y + 2)2 =16
(C) (x – 4)2 + (y – 4)2 = 8
(D) (x – 2)2 + (y – 2)2 =12
Answer: (A) Image of P(3, 5) on the line x – y + 1 = 0 is
(x-3)/1 = (y-5)/-1 = (-2(3-5+1))/ 2 = 1
x = 4, y = 4 Image is (4, 4) Which lies on (x – 2)2 + (y – 4)2 = 4
Q. 11: If Rolle’s theorem holds for the function f(x) = x3 – ax2 + bx – 4, x ∈ [1, 2] with f'(4/3) = 0, then ordered pair (a, b) is equal to :
(A) (–5, 8)
(B) (5, 8)
(C) (5, –8)
(D) (–5, –8)
Answer: (B) f(1) = f(2) ⇒ 1 – a + b –4 = 8 – 4a + 2b –4 3a – b = 7 …(1) f’(x) = 3x2 – 2ax + b ⇒f’(4/3) = 0 ⇒ 3 x 16/9 – 8a/3 + b =0 ⇒–8a + 3b = –16 …(2) a = 5, b = 8
Q. 12: If a curve passes through the origin and the slope of the tangent to it at any point (x, y) is (x2-4x+y+8)/(x-2), then this curve also passes through the point :
(A) (4, 5)
(B) (5, 4)
(C) (4, 4)
(D) (5, 5)
Answer: (D)
dy/dx = [(x-2)2 +y+4]/(x-2) = (x-2) + (y+4)/(x-2)
Let x – 2 = t ⇒ dx = dt and y + 4 = u ⇒dy = du dy/dx = du/dt du/dt = t + u/t ⇒ du/dt – u/t = t Here, IF = 1/t u. (1/t) = ∫ t.(1\t) dt ⇒ u/t = t + c ⇒ (y+4)/(y-2) = (x – 2) + c Passing through (0, 0) c = 0 ⇒ (y + 4) = (x – 2)2
Q. 13: The value of ∫1-1 x2 ex^3 dx where [t] denotes the greatest integer ≤ t,is :
(A) (e+1)/3
(B) (e-1)/3e
(C) (e+1)/3e
(D) 1/3e
Answer: (C)
Q. 14: When a missile is fired from a ship, the probability that it is intercepted is 1/3 and the probability that the missile hits the target, given that it is not intercepted, is 3/4. If three missiles are fired independently from the ship, then the probability that all three hit the target, is:
(A) 1/8
(B) 1/27
(C) 3/4
(D) 3/8
Answer:(A) Probability of not getting intercepted = 2/3 Probability of missile hitting target = 3/4 Probability that all 3 hit the target = (2/3 x3/4)3 = ⅛
Q. 15: If the curves, x2/a + y2/b and x2/c + y2/d intersect each other at an angle of 90°, then which of the following relations is true ?
(A) a + b = c + d
(B) a- b = c – d
(C) ab = (c+d)/(a+b)
(D) a-c = b+d
Answer: (B)
x2/a + y2/b ………(1)
diff :
2x/a +2y/b dx/dy =0
y/b . dx/dy = -x/a
dx/dy = -bx/ay……(2)
x2/c + y2/d………(3)
Diff :
dy/dx = -dx/cy ………(4)
m1m2 = –1
⇒ -bx/ay × -dx/cy =−1
⇒ bdx2 = – acy2…….(5) (1)–(3) ⇒ (1/a – 1/c)x2 + (1/b – 1/d) y2 = 0 Solve above equation using 5 ⇒ (c – a) – (d – b) = 0 ⇒ c – a = d – b ⇒ c – d = a – b
Q. 16: A tangent is drawn to the parabola y2 = 6x which is perpendicular to the line 2x + y =1. Which of the following points does NOT lie on it?
(A) (0,3)
(B) (-6,0)
(C) (4,5)
(D) (5,4)
Answer: (D) Equation of tangent : y = mx + 3/(2m) mT = (1/2) (Because perpendicular to line 2x + y = 1) Therefore, tangent is: y = x/2 + 3 ⇒ x – 2y + 6 = 0
Q. 17: Let f, g: N→N such that f(n + 1)= f(n)+ f(1) for all n ∈ N and g be any arbitrary function. Which of the following statements is NOT true ?
(A) f is one-one
(B) If fog is one-one, then g is one-one
(C) If g is onto, then fog is one-one
(D) If f is onto, then f(n) = n for all n ∈ N
Answer: (C) f(n + 1) = f(n) +f(1) ⇒ f(n + 1)- f(n)= f(1) → A.P. with common difference = f(1) General term = Tn = f(1) + (n-1)f(1) = nf(1) ⇒f(n) = nf(1) Clearly f(n) is one-one. For fog to be one-one, g must be one-one. For f to be onto, f(n) should take all the values of natural numbers. As f(x) is increasing, f(1)=1 ⇒f(n)=n If g is many one, then fog is many one. So “if g is onto, then fog is one-one” is incorrect.
Q. 18: Let the lines (2 – i)z = (2 + i)zˉ and (2 + i)z + (i – 2)zˉ– 4i = 0, (here i2 = –1) be normal to a circle C. If the line iz + zˉ+ 1 + i = 0 is tangent to this circle C, then its radius is :
(A) 3/√2
(B) 3√2
(C) 3/(2√2)
(D) 1/(2√2)
Answer: (C) (2–i)z = (2+i)zˉ
⇒(2–i)(x+ iy)=(2+i)(x–iy) ⇒2x–ix + 2iy + y=2x+ ix–2iy+y ⇒2ix –4iy=0 L1 ∶ x–2y=0 ⇒(2+i)z+(i –2)zˉ–4i=0. ⇒(2+i) (x + iy)+(i –2)(x–iy)– 4i = 0. ⇒2x +ix + 2iy – y + ix – 2x + y +2iy – 4i =0 ⇒2ix + 4iy – 4i =0 L2 ∶ x+2y–2 = 0 Solve L1 and L2 4y=2 or y=1/2 ∴ x = 1 Centre(1, 1/2) L3∶ iz + zˉ+ 1 + i = 0 ⇒i(x+iy)+x–iy+1+i=0 ⇒ix–y+x–iy+1+i=0 ⇒(x–y+1)+i(x–y+1)=0 Radius = distance from (1, 1/2) to x – y + 1 = 0 r = (1-1/2+1)/ √2 r = 3/2√2
Q. 19:
(A) 1/2
(B) 1/e
(C) 1
(D) 0
Answer: (C)
Q. 20: The total number of positive integral solutions (x, y, z) such that xyz = 24 is
(A) 36
(B) 45
(C) 24
(D) 30
Answer: (D) x.y.z = 24 x.y.z= 23 × 31 x=2(a1) ⋅3(b1) y = 2(a2) ⋅3(b2) z = 2(a3)⋅3(b3) a1, a2, a3 ∈ {0,1,2,3} b1, b2, b3 ∈ {0,1} Case 1: a1 + a2 + a3 =3 Non negative solution = (3+3-1) C(3-1) = 5C2 = 10 Case 2: b1 + b2 + b3 = 1 Non negative solution = (1+3-1) C(3-1) = 3C2 = 3 ∴ Total solutions =10×3=30
Section B
Q. 1: Let A is a matrix where x, y and z are real numbers such that x + y + z > 0 and xyz = 2. If A2 = l3, then the value of x3 + y3 + z3 is ______.
Answer: 7
Q. 2: Let A1, A2, A3,…. be squares such that for each n ≥ 1, the length of the side of An equals the length of diagonal of An+1. If the length of A1 is 12 cm, then the smallest value of n for which area of An is less than one is ____________.
Answer: (9)
Q. 3: The locus of the point of intersection of the lines (√3)kx + ky – 4√3 = 0 and √3x – y – 4√3 k = 0 is a conic, whose eccentricity is _________.
Q. 5: The graphs of sine and cosine functions, intersect each other at a number of points and between two consecutive points of intersection, the two graphs enclose the same area A. Then A4 is equal to ___________
Answer: (64)
Q. 6: The number of points, at which the function f(x) = |2x + 1| – 3|x+2|+|x2 + x–2|, x ∈ R is not differentiable, is ________.
Answer: 2 Check at 1, –2 and -1/2 Non-Differentiable at x = 1 and -1/2
Q. 7: If the system of equations
kx + y + 2z = 1
3x – y – 2z = 2
–2x – 2y – 4z = 3
has infinitely many solutions, then k is equal to __________.
Q. 8: The total number of numbers, lying between 100 and 1000 that can be formed with the digits 1, 2, 3, 4,5, if the repetition of digits is not allowed and numbers are divisible by either 3 or 5 is ________.
Answer: 32
Q. 9: Let f(x) be a polynomial of degree 6 in x, in which the coefficient of x6 is unity and it has extrema at x = –1 and x = 1. If limx→0f(x)/x3 =1, then 5.f(2) is equal to _________.
A) During a pandemic years when teaching moved online, only 1 in 5 Indian schools had internet facilities
In the academic year that ended with school closures due to Covid-19, only 22% of schools in India had internet facilities, according to Education Ministry data released on Thursday. Among government schools, less than 12% had internet in 2019-20, while less than 30% had functional computer facilities. This affected the kind of digital education options available to schools during the pandemic, as well as the plans for hybrid learning in the days ahead. The Unified District Information System for Education Plus (UDISE+) report collates data from more than 15 lakh schools across the country. As the first wave of Covid-19 entered India in early 2020, schools were closed in mid-March, just weeks before the end of the 2019-20 academic year. The vast majority of the country’s 26 crore school children have not stepped foot in a school since then, depending on various forms of distance education instead. The availability of digital education whether via live, synchronous teaching on apps like Zoom, or through recorded lectures, emails, Whatsapp or educational apps was largely dependent on whether schools, teachers and parents had access to the necessary infrastructure. In many States, teachers came to school and taught in their own empty classrooms, using their blackboards and lab facilities, while facing a computer screen that communicated the lessons to their students at home. However, the UDISE+ data makes clear the digital divide which made this a viable option only in some States. In many Union Territories, as well as in the State of Kerala, more than 90% of schools, both government and private, had access to working computers. In States such as Chhattisgarh (83%) and Jharkhand (73%), installation of computer facilities in most government schools paid off, while in others such as Tamil Nadu (77%), Gujarat (74%) and Maharashtra (71%), private schools had higher levels of computer availability than in government schools. However, in States such as Assam (13%), Madhya Pradesh (13%), Bihar (14%), West Bengal (14%), Tripura (15%) and Uttar Pradesh (18%), less than one in five schools had working computers. The situation is worse in government schools, with less than 5% of UP’s government schools having the facility. The connectivity divide is even starker. Only three States — Kerala (88%), Delhi (86%) and Gujarat (71%) — have internet facilities in more than half their schools. This will make it hard for most schools to implement the options for hybrid learning as schools try to re-open with staggered attendance post the pandemic.
B) Akhil Gogoi set free, slams govt for ‘misusing’ NIA, UAPA
Assam MLA and Raijor Dal president Akhil Gogoi walked out of jail on Thursday after a special court of the National Investigation Agency (NIA) acquitted him of all charges in the second of the two cases related to the anti-Citizenship (Amendment) Act violence in the State in December 2019. The activist-turned-politician, who represents the Sibsagar Assembly constituency in eastern Assam, was discharged in the first case on June 25. He was in jail since December 2019 and had spent the last few months in the Gauhati Medical College and Hospital due to ailments. The NIA had pursued the cases filed under relevant sections of the Indian Penal Code and the Unlawful Activities (Prevention) Amendment Act (UAPA) for alleged criminal conspiracy, sedition, promoting enmity between groups on grounds such as religion, race and language, assertions against national integrity, and support to a terrorist organisation. The cases were registered on the basis of two FIRs lodged with the Chabua police station in Dibrugarh district and the Chandmari police station in Guwahati. The Chandmari case had the additional charge that he had links with a Maoist organisation. The court has proven that the government’s charges against me were bogus. This is a historic day for the Indian judiciary, Gogoi told journalists after his release. A case was filed against me for my links with CPI [Maoist]. In another case, I was charged with inciting violence during the anti-CAA agitation. The court proved these were fake charges that kept me in jail for more than a year and a half, he said. He slammed the BJP-led State government for misusing the UAPA and NIA. The court’s judgement would influence future cases, he stated.
C) A journalist is not expected to dramatise incident and make news: Allahabad HC
A journalist is not expected to dramatise a sensational and horrifying incident and make news by putting his subject in a pitiable condition and in danger of death, the Allahabad High Court has noted. The court made the observation while rejecting the bail plea of a journalist in Lucknow, who is co-accused of abetting a person’s suicide and filming it outside the U.P. Vidhan Sabha last year. The journalist keeps an eye on anticipated or sudden events happening in society and brings them to the information of all the people through various news media without any tampering, this is his business, the court said on June 21. Journalists Shamim Ahmad and Naushad Ahmad allegedly contacted a man who was facing eviction from his house and induced him to set himself ablaze in front of the State Assembly so that they may film the incident and telecast it on television. The journalists allegedly surmised that if the tenant of the house did as they suggested no one would be able to force him to evict from the house. The man allegedly poured oil and set himself on fire, while the accused scribes were filming it. The person died in a hospital on October 24, 2020. An FIR was lodged in the matter under various sections of the Indian Penal Code. Justice Vikas Kunvar Srivastav said that from the details and statements recorded in the case, it was prima facie established that the accused Shamim Ahmad tempted the deceased that if he attempted to take his own life in front of the U.P. Assembly, the video-filmed act would prevent him from being evicted.
INTERNATIONAL NEWS
A) Trump Organisation CFO surrenders ahead of expected charges Trump Organisation CFO Allen Weisselberg surrendered to authorities early on Thursday ahead of expected charges against him and former President Donald Trump’s company, according to multiple news outlets. Weisselberg was seen walking into the courthouse in lower Manhattan around 6:20 a.m. with his lawyer. New York prosecutors are expected to announce the first criminal indictment on Thursday in a two-year investigation into Trump’s business practices, accusing his namesake company and Weisselberg of tax crimes related to fringe benefits for employees. The charges against the Trump Organisation and its chief financial officer, Allen Weisselberg, remained sealed on Wednesday night but were to be unveiled ahead of an afternoon arraignment at a state court in Manhattan, according to two people familiar with the matter.
B) Nine European countries authorise Covishield.
At least nine countries in Europe have given recognition to the Covishield vaccine produced by the Serum Institute of India (SII), informed sources have said. The confirmation came on Thursday when the European Union started the ‘Green Pass’ facility, which will allow travellers vaccinated with an authorised set of vaccines to travel within the EU zone covering 27 countries. The recognition by the nine countries is a ‘national’ move by the states and not by the European Union headquartered in Brussels. The list of EU Member States that have recognised Covishield as a valid vaccine includes Austria, Germany, Slovenia, Greece, Iceland, Ireland, Spain. That apart, Estonia has confirmed that it will recognise all the vaccines authorised by the Government of India for travel of Indians to Estonia, said an informed source. Switzerland, not a EU member, too allows Covishield as Schengen state. Supporting the move by Germany, ambassador Walter J. Lindner said, Confirming that a double shot of Covishield is fully recognised by Germany as valid proof of anti-COVID vaccination. Germany however has a travel ban in place for Indians as India has been recognised as a virus variant country. This (confirmation) does nevertheless not modify existing travel or visa restrictions for travellers from areas of concern/ virus variant areas, said Lindner, clarifying the German position on travel from India. The clearance by the nine European countries came on a day when the ‘Green Pass’ was introduced by the EU which is meant for travel within 27 Member States of the European Union. The clearance by the nine countries, however, is unlikely to translate into an automatic clearance of Covishield as an equivalent to the ‘Green Pass’, which recognises Pfizer/BioNTech’s Comirnaty, Moderna’s Spikevax, Oxford-AstraZeneca’s Vaxzevria and Johnson & Johnson’s Janssen. The European Medicines Agency (EMA) that created the authorised list is yet to include Covishield, which is based on AstraZeneca formula, as an acceptable vaccine. The clearance of Covishield by nine European countries showed that some EU Member States are making individual policies that are suitable to their health and international requirements. Air France, for example, has declared that Indian passengers who have taken doses cleared by EMA (Pfizer-BioNTech’s Comirnaty, Moderna’s Spikevax, Vaxzevria of Oxford-AstraZeneca and Janssen by Johnson & Johnson) are required to undergo 7 days of ‘mandatory quarantine’. These passengers must present a Covid-19 vaccination certificate demonstrating that at least 4 weeks have elapsed since the first dose of Johnson & Johnson vaccine, or 2 weeks have elapsed since the second dose of the Pfizer-BioNTech, AstraZeneca or Moderna vaccine, or since the first dose of these vaccines if the passenger holds proof that they previously have been infected with Covid-19, said Air Frace laying down the domestic requirement which places additional conditions on even those Indian passengers who have taken doses of the vaccines meant for the ‘Green Pass’ from the EU.
A) Covid-19 compensation: SC pulls up NDMA for ‘failing to perform its duty’
The Supreme Court on Wednesday pulled up the National Disaster Management Authority (NDMA), of which the Prime Minister is the ex officio chairperson, for failing to perform its duty to recommend ex gratia assistance for families of those who lost their loved ones to the Covid-19 pandemic. The National Authority [NDMA] failed to perform its duty, the Supreme Court held. A Bench led by Justice Ashok Bhushan gave the NDMA six weeks to frame the guidelines for fixing the ex gratia meant for these families. The court, however, left it to the wisdom of the NDMA to fix the amount of ex gratia. The PIL petitioners, advocate Gaurav Kumar Bansal and Reepak Kansal, had asked for a payment of ₹4 lakh each to the families of the Covid-19 dead. Not proper for the court to ask government to pay a particular amount, the court said. The judgment, pronounced by Justice M.R. Shah on the Bench, held that the government could not excuse itself of its duty to pay ex gratia by saying that such payments would entail huge expenditure. The court pointed to Section 12 of the Disaster Management Act of 2005 and said the term ‘minimum standards of relief’ mentioned in the provision included payment of ex gratia. Justice Shah dismissed Solicitor General Tushar Mehta’s submission that Section 12 was merely recommendatory and not mandatory. Instead, the court drew the government’s attention to the word shall used in Section 12 and said this made the payment of ex gratia to victims’ families a mandatory and statutory duty. The court, in this regard, said Section 12 (iii) held that the National Authority shall recommend guidelines for the minimum standards of relief to be provided to persons affected by disaster, which shall include ex gratia assistance on account of loss of life. The court, meanwhile, directed that death certificates of Covid-19 patients should show the exact cause of death. As regards insurance and social security claims of those affected by Covid-19, the court ordered the Union of India to take appropriate steps in compliance with the 15th Finance Commission Report.
B) Bharat Biotech denies any wrongdoing over Brazil Covaxin deal
Covaxin manufacturer Bharat Biotech on June 30 asserted that it had followed a routine ‘step-by-step approach’ towards getting contracts and regulatory approvals for supply of the vaccine in Brazil and added that it had neither received any advance payments nor supplied any vaccines to the South American country’s Health Ministry. Reacting to media reports, including a Rio De Janeiro-datelined news report in which Reuters cited Brazil’s Health Minister Marcelo Queiroga announcing the suspension of a $324-million Indian vaccine contract in the wake of allegations of irregularity linked to President Jair Bolsonaro, the Hyderabad-based company laid out the timeline of developments related to its efforts to supply the vaccine to Brazil. Stating that discussions with the Brazilian Ministry had begun in November 2020, Bharat Biotech said that over the course of eight months, a step-by-step approach, similar to what was observed in other countries where the company sought approvals, had been followed. Subsequently, Covaxin received emergency use authorisation in Brazil on June 4. On the issue of pricing, Bharat Biotech said it had been clearly established that the vaccine would be priced between $15 and $20 per dose for supplies to governments outside India. The pricing for Brazil had also been indicated at $15 per dose. Also, while advance payments had been received from several countries at the same price points, with supplies in process, pending approvals, as of June 29, Bharat Biotech has not received any advance payments or supplied any vaccines to MOH Brazil, the company said. Elaborating on the procurement process in Brazil and other countries, Bharat Biotech said a common approach, widely accepted and established in the industry, was followed for Covid-19 vaccines and several vaccines for routine immunisation. The vaccine maker said it had partnered with Brazil’s Precisa Medicamentos for assistance, guidance and support with regulatory submissions, distribution, insurance and conduct of phase III clinical trials. The trials, approved recently by Brazil’s health regulator ANVISA, are to be conducted on 5,000 participants.
C) Drones, flying objects banned in J&K Rajouri district
The Rajouri district administration on Wednesday banned the sale or possession of flying objects, as at least five sorties of drones around the security installations in Jammu since Sunday, including the IAF’s technical airport recently, have the security agencies on the tenterhooks in J&K. In exercise of powers vested to me under Section 144 of the CrPC, a ban is imposed on storage, sale, possession, use and transport of any drone or small flying objects, toys in the district, Deputy Commissioner Rajesh Kumar Savan said in an order. Security personnel keep vigil during a high alert in Jammu on June 28, 2021 after two drones were noticed hovering over Ratnuchak-Kaluchak military station area the previous night. The order has directed all those in the district who already possess any such gadgets to deposit them with the local police station against proper receipt. The government agencies using drones for mapping, surveys and surveillance will get the local police station incharge and the executive magistrate informed. It warned of punitive action against those who violate the order. Rajouri is close to the Line of Control (LoC) in the Pir Panjal valley. The move comes two days after two explosions, apparently ferried and dropped by two drones, rocked the technical airport of the IAF in Jammu on Sunday.
D) Navjot Singh Sidhu meets Priyanka Gandhi
Former Punjab Minister Navjot Singh Sidhu, who has been leading a public battle against Chief Minister Amarinder Singh, on Wednesday had a closed-door meeting with Congress general secretary Priyanka Gandhi Vadra. The two are said to have discussed Sidhu’s role in the revamp exercise of the Punjab unit ahead of next year’s Assembly election. Post her meeting, Vadra called on former Congress president Rahul Gandhi at the latter’s house alone. Though the cricketer-turned-politician did not accompany her, Vadra is believed to have played the role of an emissary. Had a long meeting with @priyankagandhi Ji, Sidhu tweeted, along with a photograph of his with Vadra. Party insiders say Gandhi may meet Sidhu but the schedule is not known yet. On Monday, Sidhu’s team had informed reporters that the Punjab Congress leader would be meeting Gandhi and Vadra in Delhi a day later. Though he arrived in Delhi from Patiala on Tuesday, there was no scheduled meeting. When reporters asked Gandhi about it, expressing surprise, he said, I don’t know what fuss you all are making. Sidhu’s claim also stood in sharp contrast with what Punjab Chief Minister Captain Amarinder Singh experienced during his Delhi visit last week. While Capt. Singh met the three-member panel of the All India Congress Committee (AICC), set up to end Punjab factionalism, the Gandhis did not meet him. For the past couple of months, Punjab Congress has been witnessing a political turmoil with several senior leaders speaking out against their own government, and Sidhu has been at the forefront. While a majority of the MLAs are learnt to have said that Capt. Amarinder was best placed to lead the Congress into next year’s Assembly elections, many complained about his inaccessibility, predominance of bureaucrats in running the affairs of the State, and the existence of liquor, sand and transport mafia. Sidhu had gone to the extent of alleging that the Chief Minister was protecting the Badals of the Shiromani Akali Dal (SAD) in most cases, including the 2015 sacrilege case in which the Guru Granth Sahib was desecrated at Bargari (Faridkot). The three member AICC panel — comprising Mallikarjun Kharge, Harish Rawat and J.P. Aggarwal — in its report is learnt to have suggested around 20 action points, including action in the sacrilege case as well as cracking down on various mafia operating in the State. Since then, a special investigation team has questioned former Chief Minister Parkash Singh Badal and Lok Sabha member Sukhbir Singh Badal, who was the Deputy CM when Akalis were in power. The AICC panel is also learnt to have suggested that Sidhu needed to be suitably accommodated in the party or the government. However, Capt. Amarinder is said to be resisting any move to make Sidhu the Pradesh Congress Committee chief. The Punjab unit of the Congress is expected to be revamped in the next few days based on the feedback party leaders gave to the AICC panel as well as Gandhi.
INTERNATIONAL NEWS
A) U.S. aircraft involved in Black Sea incident: Putin
Russian President Vladimir Putin said a U.S. reconnaissance aircraft was operating in sync with a British destroyer during last week’s Black Sea incident. Moscow said one of its warships fired warning shots and a warplane dropped bombs in the path of British destroyer Defender on Wednesday to force her out of an area near the Crimean Peninsula, which Russia had annexed in 2014. Britain denied that account, insisted its ship wasn’t fired upon and said it was sailing in Ukrainian waters. Speaking Wednesday during a marathon live call-in show, Putin said that the U.S. aircraft’s apparent mission was to monitor the Russian military’s response to the British destroyer. He added that Moscow was aware of the U.S. intentions and responded accordingly to avoid revealing sensitive data. In Wednesday’s incident, Britain insisted the Defender had been making a routine journey through an internationally recognized travel lane and remained in Ukrainian waters near Crimea. The UK, like most of the world, recognises Crimea as part of Ukraine despite the peninsula’s annexation by Russia. Russia denounced the Defender’s move as a provocation and warned that next time it could fire to hit intruding warships if they again try to test Russian military resolve.
B) Israel seeks to build wider ties across West Asia.
Israel’s new Foreign Minister said on Wednesday his landmark visit to the United Arab Emirates was just the start of a road to wider peace in West Asia, reaching out to Arab states still wary of normalising ties. Cabinet Minister to visit the Gulf state since the countries forged ties last year, also reiterated Israel’s concern about the Iran nuclear deal being negotiated in Vienna. The trip comes amid unease in Israel and Arab capitals about U.S. efforts to reenter a 2015 nuclear pact between Iran and world powers, which then President Donald Trump quit in 2018 in a move backed by Saudi Arabia and its Gulf allies. This visit isn’t the end of the road to peace, it is just the beginning, he told reporters. Our hand is out stretched. I hope this visit is the first of many and that together with our new friends we continue to make history in the entire region. The UAE and Bahrain normalised ties with Israel under so-called Abraham Accords brokered by Mr. Trump’s administration, creating a new regional dynamic based on mutual concern over Iran and Islamist groups. Sudan and Morocco followed suit. Mr. Lapid declined to comment on whether Israel was coordinating with Gulf states over Iran, saying he did not want to embarrass his hosts by remarking on a neighbour while in the UAE, but noted his country was worried about the nuclear deal. Gulf states have called for a stronger deal of longer duration that also addresses Tehran’s missiles programme and support for regional proxies that Iran has denied giving.
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Q. 1: An electron with kinetic energy K1 enters between parallel plates of a capacitor at an angle ‘α’ with the plates. It leaves the plates at angle ‘β’ with kinetic energy K2. Then the ratio of kinetic energies K1 : K2 will be:
, then
AAT = I2, then the value of α4 + β4 is:
. If the area of the parallelogram whose adjacent sides are represented by the vectors a￫ and a￫ is 8√3 square units, then 
is ______.
and 
be three given vectors. If →r is a vector such that 
and →r.→b =0, then 
