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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.
| 13451. |
At some temperature T, a bronze pin is a little large fit into a hole drilled in a steel block. The change in temperature required for an exact fit is minimum when (coefficient of linear expansion of bronze is grearter than that of steel) |
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Answer» only the BLOCK is HEATED |
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| 13452. |
(A) : Wooden block floats in a liquid and some fraction of its volume is immersed in the liquid. If the vessel is accelerated upwards with some acceleration then the fraction of volume immersed remains constant. (R ) : The block experience a net upward force which can be provided by extra up thrust. So the volume immersed will increase. |
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Answer» Both (A) and (R ) are TRUE and (R ) is the correct EXPLANATION of (A) |
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| 13453. |
A car is moving along a straight line OP as shown in the figure. lt moves from O to P in 18s and returns from P to Q 6s. Which of the following statements not correct regarding the motion of the car A: The average speed of the car in going from O to P and come back to Q is 20ms""^(-1). B: The average velocity of the car in going from O to P and come back to 0 is 15ms""^(-1) |
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Answer» A,B are TRUE |
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| 13454. |
A cycle followed by an engine (made of one mole of perfect gas in a cylinder with a piston) is shown in figure. A to B: volume constant B to C: adiabatic C to D: volume constant D to A: adiabatic V_C=V_D=2V_A=2V_B (a) In which part of the cycle heat is supplied to the engine from outside ? (b) In which part of the cycle heat is being given to the surrounding by the engine ? (c) What is the work done by the engine in one cycle ? Write your answer in term of P_A,P_B,V_A(d) What is the efficiency of the engine ? (gamma=5/3 for the gas , C_V=3/2R for one mole ) |
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Answer» Solution :(a) For the process A to B, volume is constant So, dV = 0 `therefore`dW = 0 From FIRST LAW of thermodynamics in, dQ = dU + đW, dW = 0 `therefore` dQ=dU Hence, in this process heat supplied is utilised to increases internal energy of the system. As pressure is increased at constant volume (From Gay-Lussac.s law) i.e. P`prop` T and internal energy is proportional to TEMPERATURE and hence internal energy increases with temperature `therefore dU gt 0` `therefore` From dQ=dU, dQ `gt` 0 , so in process AB heat is supplied to the system. (b)From dQ = dU, dQ > 0, so in process AB heat is supplied to the system. (b) In CD process, volume remains constant but pressure decreases. Hence, from P`prop`T, temperature also decreases so heat is being GIVEN to the surrounding by the engine. (c) To calculate work done by the engine in one cycle , we should calculate work done in each part separately. `therefore W_(AB)=int_A^P P dV=0` ...(1) `W_(CD)=int_(V_C)^(V_D) P dV=0`...(2) `W_(BC)=int_(V_B)^(V_C) P dV=K int_(V_B)^(V_C) (dV)/V^gamma=K/(1-gamma)[V^(1-gamma)]_(V_B)^(V_C)` `=1/(1-gamma)[PV]_(V_B)^(V_C)` (adiabatic) `=(P_C V_C -P_B V_B)/(1-gamma)`...(3) Similarly , `W_(DA)=(P_AV_A-P_DV_D)/(1-gamma)` (adiabatic) ...(4) B and C are along adiabatic curve BC, `therefore P_B V_B^gamma =P_C V_C^gamma` `therefore P_C=P_B(V_B/V_C)^gamma=P_B(1/2)^gamma =2^(-gamma)P_B`...(5) Similarly, `P_D=2^(-gamma)P_A` ...(6) Total work done by the engine in one cycle ABCDA `W=W_(AB)+W_(BC)+W_(CD)+W_(DA)` `=0+W_(BC)+0+W_(DA)` `W=1/(1-gamma)(P_C V_C-P_B V_B)+1/(1-gamma)(P_AV_A-P_DV_D)` `W=1/(1-gamma)[2^(-gamma)P_B(2V_B)-P_BV_B]+1/(1-gamma)[P_A V_A-2^(-gamma)P_A (2V_A)]` `W=1/(1-gamma)[P_B V_B 2^(-gamma+1) - P_B V_B +P_A V_A -2^(-gamma+1)P_A V_A]` `therefore W=1/(1-gamma) [P_B V_B (2^(gamma-1)-1) -P_A V_A (2^(1-gamma)-1)]` but `V_A=V_B` `therefore W=(2^(1-gamma)-1)/(1-gamma)[P_B V_A-P_A V_A]` `=(2^(1-gamma)-1)/(1-gamma)V_A(P_B-P_A)` Putting `gamma=5/3`and simplifying `W=3/2 [1-(1/2)^(2/3)](P_B-P_A)V_A` (D) Efficiency `eta="Net work"/"Heat supplied"` `=(3/2[1-(1/2)^(2/3)](P_B-P_A)V_A)/(3/2(P_B-P_A)V_A)` `=[1-(1/2)^(2/3)]` |
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| 13455. |
A block of mass m hangs on a vertical spring. Initially the spring is unstretched, it is now allowed to fall from rest. Find the distance the block falls (a) if the block is released slowly, (b) if the block is released suddenly. |
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Answer» |
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| 13456. |
In a two block system in figure match the following. {:("Column1","Column2"),("A Velocity of center of mass","P Keep on changing all the time"),("B Momentum of center of mass","Q First decreases then become zero"),("C Momentum of 1kg block","R Zero"),("D Kinetic energy of 2kg block","S Constant"):} |
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Answer» SIMILARLY, `p_(CM)` =0` NET force on the system is zero, hence `V_(CM)` and `p_(CM)` will remain constant. Velocity of 1 kg and 2kg blocks keep on changing initially and finally both of them stop as `V_(CM)` was zero. |
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| 13457. |
A number of bullets are fired horizontally with different velocities from the top of a water they reach the ground |
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Answer» at same TIME with same velocity |
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| 13458. |
From the top of a tower of height 100m a 10gm block is dropped freely and a 6gm bullet is fired vertically upwards from the foot of the tower with velocity 100ms^(-1) simultaneously. They collide and stick together. What is the common velocity after the collision? (g=10ms^(-2)) |
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Answer» |
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| 13459. |
If earth contracts such that its radius becomes half without change in mass, then wieght on earth becomes …... times. |
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Answer» `gprop1/R^2,R'=R/2` `(g')/g=R^2/(R')^2=R^2/(R/2)^2=4` `:.mg'=4mg` hence WEIGHT BECOMES 4 yimes. |
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| 13461. |
When the kinetic energy of the body executing SHM is (1)/(3) of potential energy, the displacement is………….% of amplitude. |
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Answer» `33` `=(1)/(2)m omega^(2) (A^(2)-x^(2))"""………"(1)` Potential energy `=(1)/(2) m omega^(2) x^(2)"""…….."(2)` Now from (1) and (2) `(1)/(2) m omega^(2) (A^(2)-x^(2))= (1)/(3)xx(1)/(2) m omega^(2) x^(2)` `therefore A^(2) -x^(2)= (x^2)/(3)` `therefore A^(2)= (4x^2)/(3)` `therefore x^(2) = (3)/(4) A^(2)` Here `x= (A)/(2)sqrt(3) = 0.866 A` = 87% amplitude. |
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| 13462. |
the dimensional formula for the modulus of elasticity is same as that for |
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Answer» strain |
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| 13463. |
Assertion : For an ideal gaas at constant temperature, the product of the pressure and volume is a constant. Reason : The mean square velocity of the molecules is inversely proportional to mass. Select the correct option from the following. |
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Answer» Both assertion and REASON are true and reason is the correct explanation of the assertion. |
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| 13464. |
Tanks A and B open at the top contain two different liquids upto certain height in them. A hole is made to the wall of each tank at a depth 'h' from the surface of the liquid. The area of the hole in A is twice that of in B. If the liquid mass flux through each hole is equal, then the ratio of the densities of the liquid respectively, is |
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Answer» `(2)/(1)` |
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| 13465. |
A stone is fallen freely from a tower and after n seconds another stone is thrown upwards with u m/s, then at what distance from top of tower second stone will overtake first stone ? |
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Answer» Solution :Suppose after t time of free FALL of first stone, second stone will meet first stone. `therefore` Distance covered by first stone in .t. time, `d _(1) = (1)/(2) g t ^(2) ""…(1)` and second stone will meet at (t-n) second to first stone. `therefore` Distance covered by second stone in `(t-n)` SECONDS, `d _(2) = u (t -n) + 1/2 g (t -n) ^(2)` `therefore` The distance covered by first stone in t s = The distance covered by second stone in `(t-n)` s `1/2 g t^(2) = u (t-n) + 1/2 g (t -n)^(2)` `therefore 1/2 g t ^(2) = u (t-n) + 1/2 g (t ^(2) - 2tn + n^(2))` `therefore 1/2 g t ^(2) = UT - un + 1/2 g t ^(2) - gtn + 1/2 gn ^(2)` `therefore 0= ut - g TN + 1/2 gn ^(2) - un` `therefore g t n- ut = 1/2 gn ^(2) - un` `therefore t (gn - u) =n ((1)/(2) gn - u )` `therefore t = (n ((1)/(2) gn -u ))/( gn - u )` Distance h covered by first stone in t time, then ` h = 1/2 g t ^(2)` `therefore h = 1/2 g [ (n ((1)/(2)gn - u ))/( gn - u ) ]^(2)` From the top of the tower, at h distance first stone will overtake the second stone. `therefore ` Distance from top of the tower, `1/2 [ (n ((1)/(2) gn - u ))/( gn - u ) ] ^(2)` |
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| 13466. |
Show that in the case of an elastic collision between a particle of mass m_(1) with a particle of mass m_(2) initially at rest , (a) the maximum angle through which m_(1) can be deflected is given by 1 - cos^(2) theta_(m) = (m_(2)^(2))/(m_(1)^(2)) when m_(1) gt m_(2) (b) theta_(1) + theta_(2) = (pi)/(2) when m_(1) = m_(2) |
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| 13467. |
A square lead slab of side 50 cm and thickness 10.0 cm is subjected to a shearing force (on its narrow face) of magnitude 9.0xx10^(4)N. The lower edge is riveted to the floor. How much is the upper edge displaced, if the shear modulus of lead is 5.6xx10^(9) Pa? |
Answer» Solution : Here `L=50cm=50xx10^(-2)m,G=5.6xx10^(9)Pa,F=9.0xx10^(4)N`. Area of the FACE on which force is applied, `A=50xx10=500sq.cm=0.05m^(2)` If `DELTAL` is the displacement of the UPPER edge of the SLAB due to tangential force F applied, then `G=(F//A)/(DeltaL//L)` (or)`DeltaL=(FL)/(GA)` `=(9xx10^(4)xx50xx10^(-2))/(5.6xx10^(9)xx0.05)=1.6xx10^(-4)m` |
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| 13468. |
A particle is observed from two frames S_(1) and S_(2). The frame S_(2) moves with respect to S_(1) with an acceleration alpha. Let F_(1) and F_(2) be the pseudo forces on the particle when seen from S_(1) and S_(2) respectively. Which of the following are not possible ? |
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Answer» `F_(1)=0, F_(2)NE 0` |
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| 13469. |
The statement ''acceleration is zero if and only if the net force is zero'' is valid in |
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Answer» non-INERTIAL FRAMES |
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| 13470. |
The angular momentum of a body is defined as the product of |
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Answer» MASS and ANGULAR velocity |
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| 13471. |
The driver of a car moving towards a rocket launching pad with a speed of 6 m s^(-1) observed that the rocket is moving with speed of10 m s^(-1). The upward speed of the rocket as seen by the stationery observer is nearly |
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Answer» 4 m `s^(-1)` ` = 11 ms^(-1)` |
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| 13472. |
What is wet friction? |
| Answer» SOLUTION :The FORCE of friction between a solid and the layer of LIQUID | |
| 13473. |
K is the force constatnt of a spring . The work done in extending it from l1to l2 is |
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Answer» K(l2-l1) |
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| 13474. |
Two pendulum bobs of masses m and 2m collide elastically at the lowest point in their motion, when the centres are at the same level as shown in the figure. If both the balls are released from a height H above the lowest point, to what heights do they rise for the first time after collision ? |
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Answer» Solution :GIVEN, `m _(1) = m , m _(2) = 2m , u _(1) = sqrt (2gH) and u _(2) = sqrt (2gH)` Since, the collision is elastic, their VELOCITIES after collision are `v _(1) = ((m -2m)/( m + 2m )) (- sqrt (2 gH)) + ((4m)/( m + 2m )) sqrt (2 gH)` `= (sqrt( 2 gH))/(3) + (4 sqrt (2 gH))/( 3 ) = 5/3 sqrt (2 gH) and v _(2) = ((2m)/( m + 2m )) (- sqrt (2 gH))+ ((2m- m )/( m + 2m )) (sqrt (2gH))` `= (sqrt ( 2 gH))/( 3)- (2 sqrt ( 2gH))/( 3) =- ( sqrt (2 gH))/(3)` i.e., the velocities of the balls after the collision are as shown in fig. Therefore the heights to which the balls rise after the collision are: `h _(1) = ((v _(1) ) ^(2))/( 2g) `(using `v ^(2) - u ^(2) - 2 gh)` or `h _(1) = ((( 5)/(3) sqrt (2 gH ) ) ^(2) )/( 2g) = (25)/(9) H and h _(2) ((v _(2)) ^(2))/(2g) = (( sqrt (2gH)) ^(2))/( 2g) = (H)/(9)` Note the collision is elastic, mechanical energy of both the balls will remain conserved, or `E _(i) or E _(f)` `IMPLIES (m + 2m) gH = mgh _(1) + 2m gh _(2) implies 3 mgH =(mg) ((25)/(9) H) + (2MG) ((H)/(9))` 
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| 13475. |
An iron wire and copper wire having same length and cross-section are suspended from same roof Young.s modulus of copper is 1/3rd that of iron. Then the ratio of the weights to be added at their ends so that their ends are at the same level is |
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Answer» `1:3` |
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| 13476. |
The internal energy of ideal gas is independentof the gas in bulk but depends only on the |
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Answer» IMPURITIES |
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| 13477. |
Which of the following is the most precise device for measuring length. |
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Answer» a Vernier callipers with 20 divisions on the sliding SCALE |
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| 13478. |
A body weight 500 N on the surface of the Earth. How much would it weight half way below the surface of Earth. |
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Answer» Solution :Weighton SURFACEOF Earthmg = 500 Weight BELOWTHE surfaceof Earthat `d=R/2` From variation of .G. with depth ` g. = g(1-d/R)` ` mg. = mg [ 1 - ((R/2))/R ]= 500 [ 1- 1/2] = 500 [1/2] ` `W. = 250 N ` |
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| 13479. |
Astring of length 2L, obeying hooke's law, is stretched so that its extension is L. the sped of the tansverse wave travelling on the string is v. if the sting is further on the string will be |
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Answer» `(1)/sqrt(2)V` T can be CALCULATE by using HOOKE's law and on stretching `mu` ALSO changes. |
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| 13480. |
On what factors does the turning effect of a force depend? |
| Answer» Solution :The MAGNITUDE of the force, its direction andthe distance between the FIXED POINT and the point of application. | |
| 13481. |
A spool of wire rests on a horizontal surface as shown in figure. As the wire is pulled, the spool does not slip at contact point P. On separate trials, each one of the forces F_(1),F_(2),F_(3) and F_(4) is applied to the spool. For each one of these forces the spool |
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Answer» will ROTATE anticlockwise if `F_(1)` is applied |
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| 13482. |
What does the area under acceleration-time graph represent for any given time interval |
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Answer» Final VELOCITY |
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| 13484. |
A ball .A. of mass .m. moving along positive x-direction with kinetic energy .K. and momentum P undergoes elastic head on collision with a stationary ball B of mass .M. . After collision the ball A moves along negative X-direction with kinetic energy K/9 Final momentum of B is |
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Answer» <P>P |
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| 13485. |
A round object of mass M and radius R rolls down without slipping along an inclined plane. The fractional force: |
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Answer» DISSIPATES KINETIC ENERGY as heat |
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| 13486. |
A bird of 3 kg is flying with a constant velocity of (2hati-4hatj)m//s and another bird of 2 kg with (2hati+6hatj)m//s. Then, the velocity of centre of mass of the system of two bird ……….. Is m/s. |
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Answer» `10hati+10hatj` `=(3(2hati-4hatj)+2(2hati+6hatj))/(2+3)` `=(6hati-12hatj+4hati+12hatj)/(5)` `=(10hati+0hatj)/(5)` `=2hati+0hatj` |
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| 13487. |
Figure shows three forces applied to a trunk that moves leftward by 3 m over a smooth floor. The force magnitudes are F_(1) = 5N, F_(2) = 9N, and F_(3) = 3N. Find net work done on the trunk by the three forces. |
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Answer» |
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| 13488. |
Assertion Prssure of a gas is 2/3 times translational kinetic energy of gas molecules. ReasonTranslational degree of freedom of any type of gas is three, whether the gas is monoatomic, diatomic or polyatomic. |
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Answer» If both Assertion and Reason are correct and Reason is the correct explanation of Assertion. `THEREFORE""p=2/3(K_(T))/V=2/3.` (translational kinetic ENERGY per unit VOLUME). |
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| 13489. |
(A) : The difference in the value of acceleration due to gravity at pole and equator is proportional to square angular velocity of earth. (R) : The value of acceleration due to gravity is minimum at the equator and maximum at the pole. |
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Answer» Both (A) and (R) are true and (R) is the CORRECT EXPLANATION of (A) |
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| 13490. |
Gas is taken through a cyclic process completely once. Change in the internal energy of the gas is |
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Answer» Iinfinity |
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| 13491. |
A cotton reel has a inner radius r and outer radius R. Mass of the reel is M and moment of inertia about longtudinal rotational axis is I. A force P is applied at the free end of thread wraped of the reel as shown in fig. If the reel moves without sliding. a. Determine the frictioal force exerted by the table on the reel and the direction in which it acts. b. I what direction does the reel begin to move? |
Answer» Solution :Method 1: As the body rolls, point `F` will be instantasneous centre of rotation (`ICR`). There is clockwise torque about `ICR`. Hecne centre of MASS will have acceleratioin in eright direction. As there is net torque about `ICR` in clockwise sense. Hence the reel will rotate in clockwise direction.  Hence the reel moves to the RIGHT. Force `P` exerts ANTICLOCKWISE torque, therefore FRICTION force must exert a larger clockwise torque to produce clockwise rolling. The equations of the motion are `SigmaF_(x)=F-f=Ma`.....i `SigmaF_(y)=N-Mg=0`............ii `Sigmatau=fxxR-Fxxr=Ialpha`..............iii As the reel rolls without slipping `a=Ralpha` From eqn i and iii we get `(F-f)/M=(R(fxxR-Fxxr))/I` `f=(F(I+MRr))/(I+MR^(2))` Force of friction comes out to be positive, thereforeour assumption about the direction of friction force was correct. On substitutingthe expression for in Eqn i we obtain `a=(FR(R-r))/(I+MR^(2))` As `Rgtr`, a positive and towards right. Similarly, from eqn iii, we obtain `alpha=(F(R-r))/(I+MR^(2))` Which is positive i.e., net torque on the reel is clockwise. Method 2: We can apply torque equation about.  ICR `implies F(R-r)=I_(p)alpha` `F(R-r)=[I+MR^2]alpha` `implies alpha=(F(R-r))/([I+MR^(2)])` |
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| 13492. |
Two similarspringsP and Q have springconstant k_(p) and k_(Q) such that k_(p) gt k_(Q) . They are stretched , first by the same amount (case a,) then by the same force (case b ) . The work done by the springs W_(P) and W_(Q) are related as , in case (a) and case (b) , respectively : |
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Answer» `W_(P) =W_(Q),W_(P)gt W_(Q)` `(W_(P))/(W_(Q))=(1/2k_(P)x^(2))/(1/2K_(Q)x^(2))=(K_(P))/(K_(Q))` But `K_(P) gt K_(Q)` is given , ` :. W_(P) gt W_(Q)` For Case (b) : `F_(1) = F_(2) = F ` For constant FORCE `W = (F^(2))/(2K):. W prop 1/K ` ` :. (W_(P))/(W_(Q)) = (K_(Q))/(K_(P))` but , `K_(P) gt K_(Q) rArr W_(P) lt W_(Q)` ` :. W_(Q) gt W_(P)` . |
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| 13493. |
A massless piston which can slide inside a smooth vertical cylinder is attached with a light spring initially in its natural length as shown with force constant k. Now, a mass 'm' is placed on the piston. 'A' is cross-sectional area of the cylinder and the piston. Find the work done by the gas as it expands, if the piston moves up by a distance x (consider P_0 as the atmospheric pressure). |
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Answer» Solution :As the PISTON is massless, net force on it at every instant is zero. `=`upward force = downward force `=PA = kx +mg + P_0A` `(or)P=P_0+(kx) /(A) +(mg)/(A ) ,dW =PDV =P( Adx) = (AP_0+kx +mg) dx ` ` thereforeW= int_(0)^(x)PdV(or)W= int_(0)^(x) (AP_0+kx +mg) dx=P_0Ax+1/2kx^2+mgx` (as ` Ax = Delta V)` `(or)W=P_0 DeltaV +1/2kx^2 +mgx` The result can be STATED in a DIFFERENT manner as under. The gas does work against the atmospheric pressure `P_0` (which is constant), the spring force kx (which varies linearly with x) and the gravity force mg (which is again constant). ` thereforeW_1= `Work done against `P_0 =P_0 Delta V` `W_2= `work done against `kx = 1/2kx^2` and `W_3` = Work done against mg = mgx So, that `W_("TOTAL") = W_1+W_2+W_3 =P_0Delta V +1/2 kx^2+ mgx` 
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| 13495. |
Check the correctness of the equation 1/2mv^(2)=mgh, using dimensions. |
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Answer» SOLUTION :The DIMENSIONAL formula of `1/2mv^(2)` is `M[LT^(-1)]^(2)=[ML^(2)T^(-2)]` and that of mgh is `[MLT^(-2)L]=[ML^(2)T^(-2)]` Each term on both SIDES of the EQUATION has the same dimensions. So the equation is dimensionallly correct. |
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| 13496. |
At a distance of 3cm from the mean position, the acceleration of a paticle performing S.H.M is 12cm//sec^(2). the peridic time of the particle is |
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Answer» `3.14 SEC` |
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| 13497. |
A boat of area 10 m^2 floating on the surface of a river is made to move horizontally with a constant speed2m/sby applying a tangential force. If the river is 1m deep and the water in contact with the bed is stationary , find the tangential force needed eta_(H_2 O) = 0.01poise |
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Answer» Solution :Velocity gradient `=(2m//s -0)/(1m-0) = 2s^(-1)` , from , Newton.s law of viscous force, `F_v = eta_A (dV)/(dy) = (10^(-2) xx 10^(-1) SI) (10m^2) (2s^(-1)) = 0.02 N` To keep the boat moving with CONSTANT velocity , net force on it must be made zero. Therefore, we need to APPLY 0.02 N in the DIRECTION of motion. |
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| 13498. |
Define two dimensional motion . Give examples . |
| Answer» Solution :If a particle is moving along a CURVED path in a PLANE, then it is said to be in TWO dimensional motion Eg Motion of a COIN on a carrom board. | |
| 13499. |
Speed of sound in a gas in proportional to ……… |
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Answer» square ROOT of ISOTHERMAL ELASTICITY |
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| 13500. |
A man can swim with a speed of 4 "km h"^(-1)in still water. How lonu does he lake lo Cross a river 1 km wide if the 0.81 in river flows steadily at 3 "km h"^(-1)and he makes his strokes normal to the river current? |
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Answer» 5 min `t = ("width of river")/("SPEED of man") = (1 km)/(4 km h^(-1))=1/4h = 15 min` |
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