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This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 98351. |
_____ does Andy’s little brother play with? A) Who B) Why C) How old D) Where |
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Answer» Correct option is A) Who |
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| 98352. |
DIRECTION Sainath enterprises is a partnership business with Amar, Akbar and Anthony as partners engaged in the production and sales of home appliances. Their capital contributions were Rs 50,00,000, Rs 50,00,000 and Rs 80,00,000 respectively with the profit sharing ratio of 5:5:8.As they are now looking forward to expanding their business it was decided that they would bring in sufficient cash to double the respective capitals. This was duly followed by Amar and Akbar but due to unavoidable reasons Anthony could not do so and ultimately it was agreed that to bridge the shortfall in the required capital a new partner should be admitted who would bring in the amount that Anthony could not bring and that partner would get share of profits equal to half of Anthony’s shares which would be sacrificed by Anthony only.Consequent to this agreement, Mahesh was admitted and he bought in the required capital and Rs 30,00,000 as premium for goodwill.Based on the above information you are required to answer the following questions:1. What will be the new profit sharing ratio of Amar, Akbar, Anthony and Mahesh? (a) 1:1:1:1 (b) 5:5:8:8 (c) 5:5:4:4 (d) None of the above 2. What is the amount of capital brought in by the new partner, Mahesh? (a) Rs 50,00,000 (b) Rs 80,00,000 (c) Rs 40,00,000 (d) Rs 30,00,000 3. What is the value of goodwill of the firm? (a) Rs 1,35,00,000 (b) Rs 30,00,000 (c) Rs 1,50,00,000(d) Rs 1,00,00,000 4. What will be the correct journal entry for the distribution of premium for goodwill brought in by Mahesh? (a) Mahesh capital a/c Dr 30,00,000 To Anthony’s capital a/c 30,00,000(Being ………………..)(b) Premium for goodwill a/c Dr 30,00,000To Anthony’s capital a/c 30,00,000 (Being ………………..)(c) Premium for goodwill a/c Dr 30,00,000 To Amar’s capital a/c 10,00,000 To Akbar’s capital a/c 10,00,000 To Mahesh’s capital a/c 10,00,000 (Being ……………………)(d) Premium for goodwill a/c Dr 30,00,000 To Amar’s capitals a/c 8,33,333 To Akbar’s capitals a/c 8,33,333 To Mahesh’s capital a/c 13,33,333 (Being ……………………) |
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Answer» Correct option is 1 (c) 5:5:4:4 2 (c) Rs 40,00,000 3 (a) Rs 1,35,00,000 4 (b) Premium for goodwill a/c Dr 30,00,000 To Anthony’s capital a/c 30,00,000 (Being ………………..) |
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| 98353. |
Sainath enterprises is a partnership business with Amar, Akbar and Anthony as partners engaged in the production and sales of home appliances. Their capital contributions were Rs 50,00,000, Rs 50,00,000 and Rs 80,00,000 respectively with the profit sharing ratio of 5:5:8. As they are now looking forward to expanding their business it was decided that they would bring in sufficient cash to double the respective capitals. This was duly followed by Amar and Akbar but due to unavoidable reasons Anthony could not do so and ultimately it was agreed that to bridge the shortfall in the required capital a new partner should be admitted who would bring in the amount that Anthony could not bring and that partner would get share of profits equal to half of Anthony’s shares which would be sacrificed by Anthony only. Consequent to this agreement, Mahesh was admitted and he bought in the required capital and Rs 30,00,000 as premium for goodwill. Based on the above information you are required to answer the following questions:1. What will be the new profit sharing ratio of Amar, Akbar, Anthony and Mahesh? (a) 1:1:1:1 (b) 5:5:8:8 (c) 5:5:4:4 (d) None of the above 2. What is the amount of capital brought in by the new partner, Mahesh? (a) Rs 50,00,000 (b) Rs 80,00,000 (c) Rs 40,00,000 (d) Rs 30,00,000 3. What is the value of goodwill of the firm? (a) Rs 1,35,00,000 (b) Rs 30,00,000 (c) Rs 1,50,00,000 (d) Rs 1,00,00,000 4. What will be the correct journal entry for the distribution of premium for goodwill brought in by Mahesh? (a) Mahesh capital a/c Dr 30,00,000 To Anthony’s capital a/c 30,00,000 (Being ………………..)(b) Premium for goodwill a/c Dr 30,00,000To Anthony’s capital a/c 30,00,000(Being ………………..)(c) Premium for goodwill a/c Dr 30,00,00030,00,000 To Amar’s capital a/c 10,00,000To Akbar’s capital a/c 10,00,000To Mahesh’s capital a/c 10,00,000(Being ……………………)(d) Premium for goodwill a/c DrTo Amar’s capitals a/c 8,33,333To Akbar’s capitals a/c 8,33,333To Mahesh’s capital a/c 13,33,333(Being……………..) |
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Answer» 1. (c) 5:5:4:4 2. (c) Rs 40,00,000 3. (a) Rs 1,35,00,000 4. (b) Premium for goodwill a/c Dr 30,00,000 To Anthony’s capital a/c 30,00,000 (Being ……………..) |
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| 98354. |
A: _____ does Natalie’s nephew do? B: He’s an architect. A) How B) Whom C) When D) What |
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Answer» Correct option is D) What |
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| 98355. |
A: _____ does Anthony go to work? B: _____ bus. A) How / By B) How well / On C) What kind / In D) How / In |
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Answer» Correct option is A) How / By |
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| 98356. |
A: _____ does it rain here? B: Mostly in summer and winter. A: _____ snow? B: In winter.A) When / Why B) Why / How C) What time / Whom D) When / What about |
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Answer» D) When / What about |
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| 98357. |
Find the words which are out of the logic list:A) singer B) composer C) leaflet D) audience |
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Answer» Correct option is C) leaflet |
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| 98358. |
A: _____ it snow a little in winter in Holland? B: Yes, it _____ . But last winter it _____ snow at all. A) Do / do / didn’t B) Does / do / doesn’t C) Does / does / didn’t D) Do / does / don’t |
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Answer» C) Does / does / didn’t |
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| 98359. |
Find the words which are out of the logic list:A) heat wave B) ice C) winter D) snow |
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Answer» Correct option is A) heat wave |
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| 98360. |
A family enters a winter vacation cabin whose walls are adiabatic, initially the interior temperature is the same as the outside temperature `(0^(@)C)`. The cabin consists of a single room of floor area `6m` by `4m` and height `3m`. The room contains a `2kW` electric heater. Assuming that the room is perfectly airlight and that all the heat from the heater is absorbed by the air, none escaping through the walls. The time after the heater is turned on, the air temperature reaches the comfort level of `21^(@)C` in `x xx100 sec`, then find `x` (Take `C_(v)=20.8 J//mol-K`) |
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Answer» `V=6xx4xx3=72m^(3)=72000L` `n=(72000)/(22.4)=3.2xx10^(3)mol` `Q=nX_(v)DeltaT=1.4xx10^(6)J` `t=Q//10=700sec=7xx100sec` |
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| 98361. |
Why a given sound is louder in a hall than in the open? |
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Answer» In a hall, repeated reflections of sound take place from the walls and the ceiling. These reflected sounds mix with original sound which results in increase the intensity of sound. But in open, no such a repeated reflection is possible. .’. sound will not be louder as in hall. |
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| 98362. |
A cavity is taken out from a uniform conducting sphere. Inside the cavity a dipole is placed as shown in the figure. Find the potential at point `P` (in Volt). (`q=10^(-8)C`, `l=0.1mm`, `theta=30^(@)`, `d=10 cm`, `R=12cm`) |
| Answer» Charge on the outside surface of conducting sphere is zero so field and thus potential at any point outside the sphere is zero. | |
| 98363. |
`600 J` of heat is added to a monatomic gas in a process in which the gas performs a work of `150 J`. The molar heat capacity for the process is `KR`. Then find `K`. |
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Answer» `DeltaU=600-150=450J` `DeltaU=nCDeltaT` `DeltaU=nC_(v)DeltaT` `(DeltaQ)/(DeltaU)=(C )/(C_(v))` `C=(4)/(3)C_(v)=2R` |
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| 98364. |
What are the differences between connection and conduction? |
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Answer» Conduction: Conduction is the process of direct transfer of heat through matter due to temperature difference. When two objects are in direct contact with one another, heat will be transferred from the hotter object to the colder one. The objects which allow heat to travel easily through them are called conductors. Convection: Convection is the process in which heat transfer is by actual movement of molecules in fluids such as liquids and gases. In convection, molecules move freely from one place to another. |
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| 98365. |
How does the frequency of a vibrating wire change when the attached load is immersed in water. |
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Answer» Consider a particle of mass m performing linear SHM with amilitude A. The restoring force acting on the particle is `F=-kx`, where k is the force constant and x is the displacement of the particle from its mean position. As `omega=2pif=(2pi)/(T)`, where f and T are respectively the frequency and period of SHM, `KE=(1)/(2)momega^(2)(A^(2)-x^(2))=(1)/(2)m(4pi^(2)f^(2))(A^(2)-x^(2))` `=2 m pi^(2)f^(2)(A^(2)-x^(2))=(2mpi^(2)(A^(2)-x^(2)))/(T^(2))` Thus, `KE prop A^(2) prop f^(2)` and `KE prop (1)/(T^(2))` [Note : Total energy : The total energy of a particle performing linear SHM is `E=PE+KE=(1)/(2)kx^(2)+(1)/(2)k(A^(2)-x^(2))=(1)/(2)kx^(2)+(1)/(2)kA^(2)-(1)/(2)kA^(2)` `=(1)/(2)momega^(2)=2mpi^(2)f^(2)A^(2)=(2mpi^(2))/(T^(2))A^(2)` `:. E prop A^(2), E prop f^(2), E prop (1)/(T^(2))`] |
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| 98366. |
Why two holes are made to empty an oil tin? |
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Answer» When oil comes out through a tin with one hole, the pressure inside the tin becomes less than the atmospheric pressure, soon the oil stops flowing out. When two holes are made in the tin, air keeps on entering the tin through the other hole and maintains pressure inside. |
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| 98367. |
When do the real gases obey more correctly the gas equation PV = nRT? |
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Answer» An ideal gas is one whose molecules have zero volume and no mutual force between them. At low pressure, the volume of a gas is large and so the volume occupied by the molecules is negligible in comparison to the volume of the gas. At high temperature, the molecules have large velocities and so the intermolecular force has no influence on their motion. Hence at low pressure and high temperature, the behaviour of real gases approach the ideal gas behaviour. |
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| 98368. |
A sonometer wire is stretched by a hanging metal bob. Its fundamental frequency is `n_1`. When the bob is completely immersed in water, the frequency becomes `n_2` . The relative density of the metal is |
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Answer» Consider a particle of mass m performing linear SHM with amilitude A. The restoring force acting on the particle is `F=-kx`, where k is the force constant and x is the displacement of the particle from its mean position. As `omega=2pif=(2pi)/(T)`, where f and T are respectively the frequency and period of SHM, `KE=(1)/(2)momega^(2)(A^(2)-x^(2))=(1)/(2)m(4pi^(2)f^(2))(A^(2)-x^(2))` `=2 m pi^(2)f^(2)(A^(2)-x^(2))=(2mpi^(2)(A^(2)-x^(2)))/(T^(2))` Thus, `KE prop A^(2) prop f^(2)` and `KE prop (1)/(T^(2))` [Note : Total energy : The total energy of a particle performing linear SHM is `E=PE+KE=(1)/(2)kx^(2)+(1)/(2)k(A^(2)-x^(2))=(1)/(2)kx^(2)+(1)/(2)kA^(2)-(1)/(2)kA^(2)` `=(1)/(2)momega^(2)=2mpi^(2)f^(2)A^(2)=(2mpi^(2))/(T^(2))A^(2)` `:. E prop A^(2), E prop f^(2), E prop (1)/(T^(2))`] |
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| 98369. |
The efficiency of a heat engine working between the freezing point and boiling point of water is ………(a) 6.25% (b) 20% (c) 26.8% (d) 12.5% |
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Answer» Correct answer is (c) 26.8% Freezing point of water TL = 0°C = 273K Boiling point of water TH – 100°C = 373K ∴ Efficiency, η = 1 - \(\frac{T_L}{T_H} = 1 - \frac{273}{373}\) = 0.2861; n = 26.8% |
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| 98370. |
The efficiency of an ideal heat engine working between the freezing point and boiling point of water, isA. 0.0625B. 0.2C. 0.2608D. 0.125 |
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Answer» Correct Answer - C |
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| 98371. |
What are beats ? |
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Answer» A periodic variation in loudness (intensity) due to superposition of two sound notes of slightly different frequencies sounded at the same time is called beats. The time interval between successive maxima or minima of sound at a given place is called the period of beats. The number of beats produced per unit time is called the beat frequency. Consider two sound waves of equal amplitude (A) and slightly different frequencies `n_(1)` and `n_(2)` (with `n_(1) gt n_(2)`) propagating through the same part of the medium in the same direction. These waves can be represented by the equations `y_(1)=A sin 2pi n_(1)t` and `y_(2)=A sin 2pin_(2)t` at x=0, where y denotes the displacement of the particle of the medium from its mean position. By the principle of superposition of waves, the resultant displacement of the particle of the medium at the point at which the two waves arrive simultaneously is the algebraic sum `y=y_(1)+y_(2)=A sin 2pi n_(1)t+A sin 2pi n_(2)t` Now, `sin C+sin D=2 sin ((C+D)/(2))cos((C-D)/(2))` `:. y=2A sin [2pi((n_(1)+n_(2))/(2))t]. cos[2pi((n_(1)-n_(2))/(2))t]` `=2A cos[2pi((n_(1)-n_(2))/(2))t].sin[2pi((n_(1)-n_(2))/(2))t]` Let `R=2A cos[2pi((n_(1)-n_(2))/(2))t]` and `n=(n_(1)+n_(2))/(2)` `:. y=R sin 2pi nt` The above equation shows that the resultant motion has amplitude |R| which changes periodically with time. Period of beats is the period of waxing (maximum intensity of sound) or the period of waning (minimum intensity of sound). The intensity of sound is directly proportional to the square of the amplitude of the wave. It is maximum when |R| becomes maximum `(=2A) :R=-+2A`. `:. cos[2pi((n_(1)-n_(2))/(2))t]=-+1` `:. 2pi((n_(1)-n_(2))/(2))t=0, pi,2pi,3pi`,.... `:. t=0, (1)/(n_(1)-n_(2)),(2)/(n_(1)-n_(2)),(3)/(n_(1)-n_(2))`,... `:.` Period of beats =period of waxing `=(1)/(n_(1)-n_(2))` `:.` Beat frequency `=(1)/("period of beats")=n_(1)-n_(2)` [Note : The intensity of sound is minimum (waning) when R=0. `:. cos [2pi((n_(1)-n_(2))/(2))t]=0` `:. 2pi((n_(1)-n_(2))/(2))t=(pi)/(2),(3pi)/(2),(5pi)/(2)`,....... `:. t=(1)/(2(n_(1)-n_(2))),(3)/(2(n_(1)-n_(2))),(5)/(2(n_(1)-n_(2)))`,....., `:.` Period of beats=period of waning`=(1)/(n_(1)-n_(2))` `:.` Beat frequency `=(1)/("period of beats")=n_(1)-n_(2)` Thus, waxing and waning occur alternately and the period of waning equals that of waxing.] |
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| 98372. |
Discuss analytically the composition of two linear SHMs of the same period and along the sameline. Find the resultant amplitude when the phase difference is (1)zero (2) `(pi)/(3)` (3) `(pi)/(2)`rad (4) `pi` rad |
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Answer» Let a particle be subjected to two linear SHMs of the same period, along the same line and having the same mean positon, represented by `x_(1)=A_(1) sin (omegat-alpha)` and `x_(2)=A_(2) sin (omega+beta)`, where `A_(1)` and `A_(2)` are the amplitudes and `alpha` and `beta` are the initial phases of the two SHMs. According to the principle of superposition, the resultant displacement of the particle at any instant t is the algebraic sum `x=x_(1)+x_(2)`. `:. x=A_(1) sin (oemgat+alpha)+A_(2) sin (omega+beta)` `=A_(1) sin omegat cos alpha+A_(1) cos omegat sin alpha+A_(2) sin omegat cos beta+A_(2) cos omegat sin beta` `=(A_(1) cos alpha+A_(2) cos beta) sin omegat+(A_(1) sin alpha+A_(2) sin beta) cos omegat` Let `A_(1) cos alpha+A_(2) cos beta=R cos delta" "`.....(1) and `A_(1) sin alpha+A_(2) sin beta=R sin delta" "`.......(2) Equation (3), which gives the displacement of the particle, shows that the resultant motion is also linear simple harmonic, along the same line as the SHMs superposed, with amplitude |R| and inital phase `delta` but having the same mean position and the same period as the individual SHMs. Amplitude of the resultant motion : `R^(2)=R^(2) cos^(2)delta+R^(2) sin^(2)delta` From Eqs. (1) and (2), `R^(2)=(A_(1) cos alpha+A_(2) cos beta)^(2)+(A_(1) sin alpha+A_(2) sin beta)^(2)` `=A_(1)^(2) cos^(2)alpha+A_(2)^(2) cos^(2)beta+2A_(1)A_(2) cos alpha cos beta+A_(1)^(2) sin^(2)alpha+A_(2)^(2) sin^(2) beta+2A_(2)A_(2) sin alpha sin beta` `=A_(1)^(2)(cos^(2)alpha+sin^(2)alpha)+A_(2)^(2)(cos^(2)beta+sin^(2) beta)+2A_(1)A_(2)(cos alpha cos beta+sin alpha sin beta)` `:. R^(2)=A_(1)^(2)+A_(2)^(2)+2A_(1)A_(2)cos(alpha-beta)` `:. |R|=sqrt(A_(1)^(2)+A_(2)^(2)+2A_(1)A_(2) cos (alpha-beta))" "`......(4) Initial phase of the resultant motion : From Eqs. (1) and (2), `(R sin delta)/(R cos delta)=tan delta=(A_(1) sin alpha+A_(2) sin beta)/(A_(1) cos alpha+A_(2) cos beta)` `:. tan^(-1)((A_(1) sin alpha+A_(2) sin beta)/(A_(1) cos alpha+A_(2) cos beta))" "`......(5) Now, consider Eq. (4) for |R|. Case (1) : Phase difference, `alpha-beta=0^(@)` `:. cos (alpha-beta)=1` `:. |R|=sqrt(A_(1)^(2)+A_(2)^(2)+2A_(1)A_(2))=A_(1)+A_(2)` Case (2) : Phase difference, `alpha-beta=pi//3` rad `:. cos (alpha-beta)=(1)/(2)" " :. |R|=sqrt(A_(1)^(2)+A_(2)^(2)+A_(1)A_(2))` Case (3) : Phase difference, `alpha-beta=pi//2` rad `:. Cos (alpha-beta)=0` `:. |R|=sqrt(A_(1)^(2)+A_(2)^(2))` Case (4) : Phase difference, `alpha-beta=pi` rad `:. cos (alpha-beta)=-1` `:. |R|=sqrt(A_(1)^(2)+A_(2)^(2)-2A_(1)A_(2))" " :. |R|=|A_(1)-A_(2)|` |
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| 98373. |
If the two particles performs S.H.M. of same initial phase angle but different amplitudes of individuals, then the resultant motion initial phase angle depends on |
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Answer» Let a particle be subjected to two linear SHMs of the same period, along the same line and having the same mean positon, represented by `x_(1)=A_(1) sin (omegat-alpha)` and `x_(2)=A_(2) sin (omega+beta)`, where `A_(1)` and `A_(2)` are the amplitudes and `alpha` and `beta` are the initial phases of the two SHMs. According to the principle of superposition, the resultant displacement of the particle at any instant t is the algebraic sum `x=x_(1)+x_(2)`. `:. x=A_(1) sin (oemgat+alpha)+A_(2) sin (omega+beta)` `=A_(1) sin omegat cos alpha+A_(1) cos omegat sin alpha+A_(2) sin omegat cos beta+A_(2) cos omegat sin beta` `=(A_(1) cos alpha+A_(2) cos beta) sin omegat+(A_(1) sin alpha+A_(2) sin beta) cos omegat` Let `A_(1) cos alpha+A_(2) cos beta=R cos delta" "`.....(1) and `A_(1) sin alpha+A_(2) sin beta=R sin delta" "`.......(2) Equation (3), which gives the displacement of the particle, shows that the resultant motion is also linear simple harmonic, along the same line as the SHMs superposed, with amplitude |R| and inital phase `delta` but having the same mean position and the same period as the individual SHMs. Amplitude of the resultant motion : `R^(2)=R^(2) cos^(2)delta+R^(2) sin^(2)delta` From Eqs. (1) and (2), `R^(2)=(A_(1) cos alpha+A_(2) cos beta)^(2)+(A_(1) sin alpha+A_(2) sin beta)^(2)` `=A_(1)^(2) cos^(2)alpha+A_(2)^(2) cos^(2)beta+2A_(1)A_(2) cos alpha cos beta+A_(1)^(2) sin^(2)alpha+A_(2)^(2) sin^(2) beta+2A_(2)A_(2) sin alpha sin beta` `=A_(1)^(2)(cos^(2)alpha+sin^(2)alpha)+A_(2)^(2)(cos^(2)beta+sin^(2) beta)+2A_(1)A_(2)(cos alpha cos beta+sin alpha sin beta)` `:. R^(2)=A_(1)^(2)+A_(2)^(2)+2A_(1)A_(2)cos(alpha-beta)` `:. |R|=sqrt(A_(1)^(2)+A_(2)^(2)+2A_(1)A_(2) cos (alpha-beta))" "`......(4) Initial phase of the resultant motion : From Eqs. (1) and (2), `(R sin delta)/(R cos delta)=tan delta=(A_(1) sin alpha+A_(2) sin beta)/(A_(1) cos alpha+A_(2) cos beta)` `:. tan^(-1)((A_(1) sin alpha+A_(2) sin beta)/(A_(1) cos alpha+A_(2) cos beta))" "`......(5) Now, consider Eq. (4) for |R|. Case (1) : Phase difference, `alpha-beta=0^(@)` `:. cos (alpha-beta)=1` `:. |R|=sqrt(A_(1)^(2)+A_(2)^(2)+2A_(1)A_(2))=A_(1)+A_(2)` Case (2) : Phase difference, `alpha-beta=pi//3` rad `:. cos (alpha-beta)=(1)/(2)" " :. |R|=sqrt(A_(1)^(2)+A_(2)^(2)+A_(1)A_(2))` Case (3) : Phase difference, `alpha-beta=pi//2` rad `:. Cos (alpha-beta)=0` `:. |R|=sqrt(A_(1)^(2)+A_(2)^(2))` Case (4) : Phase difference, `alpha-beta=pi` rad `:. cos (alpha-beta)=-1` `:. |R|=sqrt(A_(1)^(2)+A_(2)^(2)-2A_(1)A_(2))" " :. |R|=|A_(1)-A_(2)|` |
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| 98374. |
A radioactive sample with half-life=T emits `alpha`-particles. Its total activity is `A_(i)` at some time and `(A_(f))` at a later time. The number of `alpha`-particle emitted by the sample between these two points in time isA. `A_(i)-A_(f)`B. `(T)/(In2)(A_(i)-A_(f))`C. `(In 2)/(T)(A_(i)-A_(f))`D. `(T)/(In 2)((1)/(A_(f))-(1)/(A_(i)))` |
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Answer» Correct Answer - B |
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| 98375. |
(a). State Faraday’s law of electromagnetic induction. (b). Explain with the help of a suitable example, how we can show that Lenz’s law is a consequence of the principle of conservation of energy. |
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Answer» (a) Faraday’s laws of electromagnetic induction : (i) Whenever the amount of magnetic flux linked with a circuit changes, an emf is induced in the circuit. The induced emf lasts so long as the change in magnetic flux continues. (ii) The magnitude of emf induced in a circuit is directly proportional to the rate of change of magnetic flux linked with the circuit. (b) Lenz’s law is in accordance with the law of conservation of energy. For example : When N-pole of magnet is moved towards the coil, the upper face of the coil acquires north polarity. Therefore, Work has to be done against the force of repulsion, in bringing the magnet closer to the coil. Similarly, When N-pole of magnet is moved away, South polarity develops on the upper face of the coil. Therefore, Work has to be done against the force of attraction, in taking the magnet away from the coil. It is this mechanical work done in moving the magnet w.r.t. the coil that changes into electrical energy producing induced current. Thus, Energy is being transformed. When we do not move the magnet, Work done is zero. Therefore, Induced current is zero. Hence, Lenz’s law obeys the principle of energy conservation. Conversely, Lenz’s law can be treated as a consequence of the principle of energy conservation. |
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| 98376. |
(I) Name a device that helps to maintain a potential difference across a conductor. (II) What is meant by saying that the potential difference between two points is 1 V? (III) How much energy is given to each coulomb of charge passing through a 6 V battery? (IV) List the properties of magnetic field lines. (V) Why don’t two magnetic field lines intersect each other? |
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Answer» (i) Battery as well as a cell helps in maintaining a potential difference across a conductor. This makes the current flow. (ii) If 1 J of work is required to move a charge of amount 1 C from one point to another, then it is said that the potential difference between the two points is 1 V. (iii) The energy given to each Coulomb of charge is equal to the amount of work required to move it. Thus, we have W = V * q = 6 * 1 = 6 J (iv) Magnetic fields can be pictorially represented by magnetic field lines, the properties of which are as follows: • The field is tangent to the magnetic field lines. • Field strength is proportional to the line density. • Field lines cannot intersect • Field lines are continuous loops. • Magnetic field lines emerge from the North Pole and terminate at the South pole outside the magnet. Inside the magnet, the direction of the magnetic field lines is from the South Pole to the North Pole. (v) The magnetic field lines never intersect each other because if two or more lines intersect each other, then it means that at that point of intersection, the magnetic field has two directions at the same point. This is not possible for a magnetic field to point in more than one direction at the same point. |
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| 98377. |
Two radioactive materials `X_(1)` and `X_(2)` have decay constants `10 lamda` and `lamda` respectively. If initially they have the same number of nuclei, if the ratio of the number of nuclei of `X_(1)` to that of `X_(2)` will be `1//e` after a time `n/(9lamda)`. Find the value of `n`? |
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Answer» Correct Answer - 1 1 |
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| 98378. |
Define threshold frequency, work function and stopping potential with reference to photoelectric effect. |
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Answer» Threshold frequency : The minimum frequency of incident light for a metal surface, below which the incident light can not eject electrons, is called threshold or cut off frequency for that metal surface. Work function: The minimum amount of energy required to eject an electron from a surface without imparting any kinetic energy is called work function of the sufrace. Stopping potential : The value of negative potential of the collector corresponding to which even the fastest photo electrons are repelled back, is called stopping potential. It is also defined as the minimum value of opposing potential difference required as to just stop the photo electrons emitted with maximum kinetic energy. It is also called cut off potential. |
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| 98379. |
The device that is used to detect current in an electric circuit isa) Voltmeterb) Galvanometerc) Ammeterd) Resistor |
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Answer» b) Galvanometer |
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| 98380. |
Match the following.(Column I)(Column II)(A) Neoprene(i) Prop-2-cnal(B) Isoprene(ii) Chloroprene(C) Teflon(iii) Natural rubber(D) Acrolein(iv) Chlorofluoroethene(a) A→ (i); B→ (iii); C→ (ii); D → (iv)(b) A→ (iv); B→ (iii); C→ (i); D → (ii)(c) A→ (iii); B→ (ii); C→ (iv); D → (i)(d) A→ (ii); B→ (iii); C→ (iv); D → (i) |
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Answer» (d) A→ (ii); B→ (iii); C→ (iv); D → (i) Neoprene ⇒ Chloroprene Isoprene ⇒ Natural rubber Teflon ⇒ Chlorofluoroethene Acrolein ⇒ Prop-2-enal |
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| 98381. |
Which of the following undergoes Vulcanization? (a) Neoprene and sulphur(b) Isoprene and sulphur(c) Neoprene and styrene(d) Isoprene and styrene |
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Answer» (b) Isoprene and sulphur Vulcanization process consists of heating a mixture of natural rubber (Isoprene) with sulphur and an appropriate additive at a temperature range between 373 K to 415 K |
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| 98382. |
In a meter bridge, null point is at `l=33.7cm`, when the resistance `S` is shunted by `12Omega` resitance the null point is found to be shifted by a distance of `18.2 cm`. The value of unknown resistance `R` should be A. `13.5Omega`B. `68.8Omega`C. `3.42Omega`D. None of these |
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Answer» `(R)/(S) = (33.7)/(100 - 33.7) rArr (R)/(((12xxS)/(12+S)))=((33.7+18.2))/(100-(33.7+18.2))` solving get `R = 6.86Omega` . |
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| 98383. |
Two identical small discs each of mass m placed on a frictionless horizontal floor are connected with the help of a light spring of force constant k. The discs are also connected with two light rods each of length `2sqrt(2)m` that are pivoted to a nail driven into the floor as shown as shown in the figure by a top view of the situation. If period of small oscillations of the system is `2pi sqrt((m//k))`, find relaxed length (in meters) of the spring. |
| Answer» Correct Answer - `4.76m` | |
| 98384. |
All ammeter/voltameter are ideal. Which of the statements are correct. A. When all switch are closed, reading of `A_(1)` and `A_(3)` are same.B. When only `S_(2)` is closed than reading of `V_(2)` is E.C. When only `S_(2) S_(3)` closed then a reading of `A_(1)` is `(2E)/(3R)`D. Reading of `A_(1)` is independent on the function of switch `S_(2)`. |
| Answer» Correct Answer - B | |
| 98385. |
A milli- ammeter of range `10 mA` and resistance `9 Omega` are joined in a circuit as shown in the fig. The meter gives full scale deflection, when current in the main circuit is I, and A and D are used as terminals. The value of I is A. `1.09 A`B. 10.9 AC. zeroD. 0.109 A |
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Answer» Correct Answer - D d. `(0.1 + 0.3 + 0.6)i_1 = (9+0.9) xx 10` `i = 99 mA` `I = (99+10) mA = 0.109 A` . |
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| 98386. |
In the figure shows, all the capacitors are initially uncharged. Case I: only switch `S_1` is closed Case II: only switch `S_2` is closed Case III: both swithces are closed Find the ratio of the total energy stored in the system of capacitors for the cases I, II, and III, respectively. A. `1:2:3`B. `5:7:9`C. `6:8:9`D. `9:7:6` |
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Answer» Correct Answer - C Case I: `C_(eq)=2C,U_(1)=(1)/(2)2CE^(2)=CE^(2)` Case II: `C_(eq)=(8C)/(3),U_(2)=(1)/(2)(8C)/(3)E^(2)=(4)/(3)CE^(2)` Case III: The upper `2C` will be short-circuited. `U_(3)=(1)/(2)CE^(2)+(1)/(2)2CE^(2)=(3)/(2)CE^(2)` `U_(1):U_(2):U_(3)=6:8:9` |
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| 98387. |
Ram ans Shyam purchased two electric tea kettles A and B of same size, same thickness, and same volume of 0.4L. They studied the specification of kettles as under `Kettle A: ` specific heat capacity `= 1680 Jkg^(-1)K^(-1)` Mass = 200 g Cost = Rs. 400 Kettle B: Specific heat capacity = `2450 Jkg^(-1) K^(-1)` Mass = 400 g Cost = Rs. 400 When kettle A is switched on with constant potential source, the tea begins to boil in 6 min. When kettle B is switched on with the same source separately, then tea begins to boil in 8 min. The efficiengy of kettle is defined as (Energy used for liquid heating)/(Total energy supplied) They made discussion on specification and efficiency of kettles and subsequently prepared a list of questions to draw the conclusions. Some of them are as under (assume specific heat of tea liquid as `4200J kg^(-1)K^(-1)` and density `1000 kgm^(-3)` Efficiency of kettle A isA. `63.34%`B. `83.34%`C. `93.34%`D. `73.34%` |
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Answer» Correct Answer - B b. `eta = (0.4 xx 4200 (theta - theta_1))/(420 xx 4.8(theta-theta_1)) xx 100 = 250//3 = 83.34%` `eta = (0.4 xx 4200 (theta - theta_1))/(420 xx 6.4(theta-theta_1)) xx 100 = 62.5%` `H_(A) = 0.2 xx 1680 (theta-theta_1) + 0.4 xx 4200 (theta-theta_1)` `= 420 xx [0.8 +4] (theta - theta_1)` `H_(B) = 0.4 xx 2450 (theta-theta_1) + 0.4 xx 4200 (theta-theta_1)` `= 420 xx (2.4 + 4)(theta-theta_1)` . `H_(A)/H_(B) = 4.8/6.4 = 3/4` `H_(A) = V^2/R_(A) xx 6` `H_(B) = V^2/R_(B) xx 8` `H_(A)/H_(B) = R_(B)/R_(A) xx 3/4 but H_(A)/H_(B) = 3/4 ` Then `R_(B) = R_(A)` |
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| 98388. |
In the arrangement shown in figure when the switch `S_2` is open, the galvanometer shown no deflection for `l=L//2`. When the switch `S_2` is closed, the galvanometer shown no deflection for `l=5L//12`. The internal resistance `(r)` of `6V` cell, and the emf `E` of the other battery are respectively A. `3 Omega, 8 V`B. `2 Omega, 12V`C. `2Omega, 24 V`D. `3 Omega, 12 V` |
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Answer» Correct Answer - B b. When switch `S_1` is open `6/E = L/2//L= 1/2` `:. E = 12V` When switch `S_2` is closed, we get `(6xx10)/(10+r) = (5L)/(12L) E = 5/12 xx 12 = 5` `:. 10 + r = 12 or r = 2Omega` |
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| 98389. |
Assertion : The switch `S` shown in the figure is closed at `t = 0`. Initial current flowing through battery is `(E)/(R + r)` Reason : Initially capacitor was uncharged, so resistance offered by capacitor at `t = 0` is zero A. Statement I is true, Statement II is True, Statement II is a correct explanation for Statement I.B. Statement I is True, Statement II is True, Statement II is not a correct explanation for Statement I.C. Statement I is True, Statement II is False.D. Statement I is False, Statement II is True. |
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Answer» Correct Answer - A a. Charge on capacitor `q = CE(1-e^(-t//CR_(eq)))` `:. I = (dq)/(dt) = E/R_(eq) e^(-t//CR)` at `t = 0 , I = E/R_(eq) = E/R+r` Therefore, resistance offered by the capacitor is zero. |
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| 98390. |
Threshold wavelength of certain metal is `lamda_0` A radiation of wavelength `lamdaltlamda_0` is incident on the plate. Then, choose the correct statement from the following.A. Initially, electrons will come out from the plate.B. The ejected electrons experience retarding force due to development of positive charges on the plate.C. After some time, ejection of electrons stop.D. None of the above |
| Answer» For photoemission `lambda lt lambda_(0)` | |
| 98391. |
A frame a bed and a sliding rod `PQ` of resistance `R`, start moving with the velocities `v` and `2v` respectively parallel to a long wire carrying steady current `I`, as shown in the figure. A. charge on the capacitor at time `t` is `q=(Cmu_(0)ivln2)/(2pi)[1-e^(-t//RC)]`B. charge on the capacitor at time `t` is `q=(Cmu_(0)ivln2)/(pi)[1-e^(-t//RC)]`C. current passing through the resistor at time `t` is `i=(mu_(0)ivln2)/(2piR)[e^(-t//RC)]`D. current passing through the capcitor at time `t` is `i=(mu_(0)ivln2)/(piR)[e^(-t//RC)]` |
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Answer» `epsilon_(PC)=(vmu_(0)i)/(2pi)ln2` `epsilon_(PQ)=(2vmu_(0)i)/(2pi)ln2` `epsilon_("net")=(mu_(0)iv)/(2pi)ln2` `q=Cepsilon[1-e^(-t//RC)]` `i=(epsilon)/(R )e^(-t//RC)` |
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| 98392. |
A liquid confined inside an adiabatic is suddenly taken from state `1` to state `2` buy a single step process as shown Select correct option(s) :A. `DeltaU = -3P_(0)V_(0)`B. `DeltaH = -P_(0)V_(0)`C. `W = -3P_(0)V_(0)`D. `q = 0` |
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Answer» Correct Answer - A::B::C::D `W = DeltaU = -P_(0)(4V_(0) - V_(0)) = - 3P_(0)V_(0)` `Delta H = DeltaU + Delta(PV)` `= - 3P_(0)V_(0) + (4P_(0)V_(0)-2P_(0)V_(0)) = -P_(0)V_(0)` `q = 0` |
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| 98393. |
In the circuit shown in figure. A is a sliding contact which can move over a smooth rod PQ. Resistance per unit length of the rod PQ is `1 ohm//m`. Initially slider is just left to the point P and circuit is in the steady state. At `t=0` slider starts moving with constant velocity `v = 5 m//sec`. towards right. Current in the circuit at `t=2` sec. is A. 1 ampB. less then 1 ampC. more than 1 ampD. `(15)/(16)"amp"` |
| Answer» Correct Answer - C | |
| 98394. |
A uniform but time varying magnetic field is present in a circular region of radius R. The magnetic field is perpendicular and into the plane of the paper and the magnitude of the field is increasing at a constant rate `alpha`. There is a straight conducing rod length 2R placed as shown in the figure. The magnitude of induced emf across the rod is A. `piR^(2)alpha`B. `(piR^(2)alpha)/(2)`C. `(R^(2)alpha)/(sqrt(2))`D. `(piR^(2)alpha)/(4)` |
| Answer» Correct Answer - D | |
| 98395. |
Fig. shows a wave motion in which displacements of particles at an instant is represents as shown. Choose the correct options. A. IF the wave represents a transverse wave in a stretched string. Then an elements in regionB has max. kinetic energyB. If it represents a transverse wave in strethced string, then an elements in the region B has max potential energy.C. It it represents standing wave, point D has maximum potential energy.D. If it represents standing wave, point B has max. kinetic energy. |
| Answer» Correct Answer - ABC | |
| 98396. |
A cylindrical conductor AB of length l and area of cross-section a is connected to a battery having emf E and negligible internal resistance. The specific conductivity of cylindrical conductor varies as `sigma = sigma_(0) (1)/(sqrt(x))`, where `sigma_(0)` is constant and x is distance from end A. What is the electric field just near the end B of cylinder? A. `(E)/(l)`B. `(3E)/(2l)`C. `(2E)/(3l)`D. `(E)/(2l)` |
| Answer» Correct Answer - B::C | |
| 98397. |
A liquid of refractive index `1.6` is placed between two planoconvex identical lenses, the medium of which has refrective index 1.5. Two possible arrangements P and Q are shown. Then the system is A. Divergent in P, Convergent in QB. Convergent in P, Divergent in QC. Convergent in bothD. Divergent in both |
| Answer» Correct Answer - A | |
| 98398. |
Column -I containing equations of a particles different motions. Here `vec(v)` represents instantaneous velocity x is position along x-axis , y is positin along y-axis and t is an instants of time. Column II contains various characterstics of the motion. match appropriately and mark you answer. `{:("Column I",,"Column II"),("(A)" vec(r) =6 sin ((2pit)/5) hati,,"(P) Straight line motion"),("(B)" vec(r)=3 sin 2pit hati+3 cos 2pit hatj,,"(Q) Motion on circualr path"),("(C)" x=6 cm ((2pit)/5)","y=6 sin ((2pit)/5),,"(R) Acceleration is constant in magnitude"),(,,"(S) Velocity is not constant"):}` |
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Answer» Correct Answer - PS QRS PS |
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| 98399. |
The potential energy of a particle varies with distance `x` from a fixed origin as `U = (A sqrt(x))/( x^(2) + B)`, where `A and B` are dimensional constants , then find the dimensional formula for `AB`.A. `M^(1)L^(7//2)T^(-2)`B. `M^(1)L^(11//2)T^(-2)`C. `M^(1)L^(5//2)T^(-2)`D. None of these |
| Answer» Correct Answer - B | |
| 98400. |
Calculate the entropy change in surroundings when `1.00` mol of `H_(2)O(l)` is formed under standard conditions, `Delta_(r )H^(Θ) = -286 kJ mol^(-1)`. |
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Answer» `H_(2)(g)+1/2O_(2)(g) rarr H_(2)O(l), Delta_(f)H^(Θ)=-286 kJ mol^(-1)` This means that when `1 mol` of `H_(2)O(l)`, is formed, `286 kJ` of heat is relesed. This heat is absorbed by the surroundings, i.e., `q_(surr)=+ 286 kJ mol^(-1)` `:. DeltaS=q_(surr)/T=(286 kJ mol^(-1))/(298 K)` `=0.9597 k J K^(-1) mol^(-1)` `=959.7 J K^(-1) mol^(-1)` |
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