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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.
| 14701. |
The potential energy of a particle of mass 0.1 kg., moving along the x-axis, is given by U= 5x (x - d)J, where x is in metres. It can be concluded that: |
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Answer» the particle is ACTED upon by a CONSTANT force. |
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| 14702. |
By conversion of 1 mug mass in energy, …... energyis obtained. |
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Answer» `E=mc^2` `=10^-9xx(3xx10^8)^2` `=9xx10^7J` |
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| 14703. |
Determine the maximum ratio h//b for which the homogenous block will side without tipping under the actionof force P. The coefficient of static friction between the block and the incline is mu_(s). |
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Answer» Solution :Block has to slide WITHOUT tipping, which implies that block is in translational as well as rotational equilibrium. From conditions of equlibrium, Along SLOPING surface. `F+muN=mgsintheta` ..................i  Perpendicular to the sloping surface `N=mgcostheta` .............ii Balancing torque about point `P`, `Fxxh=mgcosthetab/2` .................iii when the block is on the verge of tipping, the normal reaction will pass through the edge of the block. from EQN i, ii and iii `F=mgsintheta-mumg cos theta`........IV On substiting `F` in eqn ii we obtain `(mgsintheta-mumgcostheta)h-mgcosthetah/2=0` Which on solving for `h//b` yields `h/b= 1/((2mu_(2)-tantheta))` |
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| 14704. |
A tap supplies water at 10^(@)c and another tap at 100^(@)c. How much hot water must be taken so that we get 20kg of water at 35^(@)c. |
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Answer» 40/9 kg |
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| 14705. |
The dimensions of the quantity namely (mu_(0)ce^(2))/(2h)where mu_(0) permeability of free space, c - velocity of light, e - electronic charge and h =(h)/(2pi) being Planck's constant |
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Answer» `[M^(0)LT]` `:. c= (1)/(sqrt(mu_(0)epsilon_(0))` or `c^(2)= (1)/(mu_(0)epsilon_(0))` `:. cmu= (1)/(cepsilon_(0)) :. (pimu_(0)ce^(2))/(h)= (pie^(2))/(cepsilon_(0)h)` `:. F= (1)/(4piepsilon_(0))(e^(2))/(r^(2))` or `(e^(2))/(r^(2))= Fe^(2)4PI :. (pie^(2))/(cepsilon_(0)h)= (Fr^(2))/(ch) xx 4pi^(2)` The dimensions of `(Fr^(2))/(ch)= ([MLT^(-2)][L^(2)])/([LT^(-1)][ML^(2)T^(-1)])= ([ML^(3)T^(-2)])/([ML^(3)T^(-2)])= [M^(0)L^(0)T^(0)]` |
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| 14706. |
Four forces of magnitudes P, 2P, 3p and 4P act along the four sides of a square ABCD in cyclic order. If the magnitude and direction of their resultant is n sqrt2P and 15^(n) find the value of n |
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| 14707. |
A rubber ball dropped from a height of 20m (g= 10 m//s^(2)) falls on the roof of an elevator going up at 2m/s. If the collision of the ball with the elevator is elastic, the ball rebounds with a velocity of |
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Answer» 20m/s |
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| 14708. |
Surface tension has the same dimensions as that of |
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Answer» coefficient of viscosity Spring constant = ` ("Force")/("Lenght")= ([MLT^(-2)])/([L])= [ML^(0)T^(-2)]` |
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| 14709. |
According to Hooke's law of elasticity if stress is increased the ratio of stress to strain. |
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Answer» increases |
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| 14710. |
A balloon is filled at 27^(@)C and 1 atm pressure by 500m^(3) He. Then find the volume of He at -3^(@)C and 0.5 mm Hg pressure. |
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Answer» <P> SOLUTION :`(P_(2)V_(2))/(T_(2))=(P_(1)V_(1))/(T_(1))``V_(2)=(P_(1)V_(1)T_(2))/(P_(2)T_(1))(1xx500xx(273-3))/(0.5xx(273+27))=(500xx270)/(0.5xx300)rArr V_(2)=900M^(3)` |
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| 14711. |
For small deformations stress is directly proportional to strain. |
| Answer» SOLUTION :for SMALL DEFORMATIONS, STRESS is directly proportional to strain. | |
| 14712. |
Two arms of a common balance are unequal. How can the balance be used to find the correct mass of an object? |
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Answer» Solution :Suppose the correct mass of the object is m and the LENGTHS of the left and right arms of the balance are X and y respectively. At first the object is kept on the left pan and the counterpoising weight `m_(1)` on the right pan Fig. HENCE mx = `m_(1)y " ""or", (m)/(m_(1))=(y)/(x) ""cdots(1)` Next the object is kept on the right pan and the counterpoising weight `m_(2)` on the left pan Fig. Hence `m_(2)x =my """or", (m_(2))/(m)=(y)/(x) ""cdots (2)` From equations (1) and (2) we GET, `(m)/(m_(1))=(m_(2))/(m) " ""or", m^(2)=m_(1)m_(2)" ""or",m = sqrt(m_(1)m_(2))` Hence, the true mass m of the object can be calculated. 
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| 14713. |
A vertical cylinder closed at both ends is fitted with a smooth piston dividing the volume into two parts each containing one mole of air. At the equilibrium temperature of 320 K, the upper and lower parts are in the ratio 4:1. The ratio will become 3:1 at a temperature of |
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Answer» 450 K or `""(mg)/A=(RT_(1))/(V_(i))-(RT_(i))/(4V_(1))=(3RT_(i))/(4V_(1))` SIMILARLY in second CASE, `""(mg)/A=(RT_(f))/(V_(2))-(RT_(f))/(3V_(2))=(3RT_(f))/(3V_(2))` Further `5V_(1)=4V_(2)` Equating EQS. (i) and (ii), we GET `""(3T_(i))/(4V_(1))=(2T_(f))/(3V_(2))` or `""T_(f)=9/8xx(V_(2))/(V_(1))xxT_(1)` `""=9/8xx5/4xx320=450K` 
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| 14714. |
Block A and B are identical having 1 kg mass each. A is tied to a spring of force constant k and B is placed in front of A (touching it). Block ‘B’ is pushed to left so as to compress the spring by0.1 m from its natural length. The system is released from this position. Coefficient of friction for both the blocks with horizontal surface ismu=0.2 (a) Take k=200/3 N//m. Kinetic energy of the system comprising of the two blocks will be maximum after travelling through a distance x0 from the initial position. Find x0. Find the contact force between the two blocks when they come to rest.(b) Take k = 100 N/m. What distance (x1) will the block travel together, after being released, before B separates from A. |
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| 14715. |
The length of the second hand of a clock is 21 cm. What will be the magnitude of linear velocity of its extreme point? |
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| 14716. |
Consider the following two equations(A) L = I omega ""(B) (dL)/(dt)=tauIn noninertial frames |
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Answer» Both A and B are TRUE |
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| 14717. |
Area under force-displacement graph is equal to |
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Answer» Impulse |
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| 14718. |
If Y=a xx b, the maximum percentager error in the measurement of Y will be |
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Answer» `((DELTA a)/(a)XX100)//((Delta B)/(b)xx100)` |
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| 14719. |
A body falling under the effect of gravity is said to be in free fall .From the given figures, identify the figure which represents uniformly accelerated motion |
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Answer»
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| 14720. |
By opening the door of a refrigerator inside a closed room: |
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Answer» you can COOL the ROOM to a certain degree |
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| 14721. |
A body of mass 2 kg is projected vertically up with a velocity of 100 ms^(-1). If it rises to a height of 400 m, the energy utilized to overcome friction is (g = 10 ms^(-2)) |
| Answer» ANSWER :D | |
| 14722. |
The sign of work done by a force on a body is important to understand. State carefully if the following quantities are positive or negative : Work done by an applied force on a body moving on a rough horizontal plane with uniform velocity. |
| Answer» SOLUTION :POSITIVE | |
| 14723. |
The sign of work done by a force on a body is important to understand. State carefully if the following quantities are positive or negative : Work done by the resistive force of air on a vibrating pendulum in bringing it to rest. |
| Answer» SOLUTION :NEGATIVE | |
| 14724. |
The sign of work done by a force on a body is important to understand. State carefully if the following quantities are positive or negative : Work done by friction on a body sliding down an inclined plane. |
| Answer» SOLUTION :NEGATIVE | |
| 14725. |
A cylinder of mass M_(c) and sphere of mass M_(s) are placed at points A and B of two inclines respectively. (See figure). If they roll on the incline without slipping such that their acceleration are the same then the ratio (sintheta_(c))/(sintheta_(s)) is : |
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Answer» `8/7` |
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| 14726. |
A block of mass 'm' is connected to one end of a spring of 'spring constant' k. The other end of the spring is fixed to a rigid support. The mass is released slowly so that the total energy of the system is then constituted by only the potential energy, then 'd' is the maximum extension of the spring. Instead, if the mass is released suddenly from the same initial position, the maximum extension of the spring now is: (g-acceleration due to gravity) |
| Answer» Answer :C | |
| 14727. |
If the Earth has no rotational motion, the weight of a person on the equator is W. Determine the speed with which the earth would have to rotate about its axis so that the person at the equator will weigh 3/4 W Radius of the Earth is 6400 km and g = 10 m//s^(2)) |
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Answer» `0.28 XX 10^(-3)` rad/s |
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| 14728. |
The piston I s now pulled out slowly and held at a diatance 2L from the top. The pressure in thecylinder between its top and the piston will then be |
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Answer» `P_(0)` |
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| 14729. |
A 5kg body is rotated in a vertical circle with a constant speed of 4ms^(-1) using a string of length 1m, when the tension in the string is 31N, then the body will be |
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Answer» at the LOWEST point |
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| 14730. |
A force vec(F) = 5i - 3j + 2k moves a particle from vec(r )_(1) = 2i + 7j + 4k " to " vec_(r )_(2) = 5i + 2j + 8k. Calculate the workdone |
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Answer» 38 units |
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| 14731. |
A monochromatic light passes through a glass slab of thickness 9cm in time t_1 IF it takes a time t_2 to travel the same distance through water . The value of (t_1-t_2) is |
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Answer» `5 TIMES 10^-11 SEC` |
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| 14732. |
A capillary tube sealed at the top has an internal radius of 0.05 cm. the tube is placed vertically in water, openend first . What should be the length of such a tube for the water column to rise in it to a length of 1 cm? Atmospheric pressure, P_(0)=1 atmosphere =1.01 xx 10^(5) Nm^(-2), and surface tension of water =70 xx 10^(-3)Nm^(-1). |
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Answer» Solution :Consider the given capillary tube of length l, RADIUS r and cross-sectional area A. Let h be the height through which the water rises in the capillary tube due to surface tension. Let R be the radius of curvature of water meniscus in the tube which is NEARLY equal to the radius of the tube because the angle of contact of water - glass surface is nearly `0^(@)`. the rising water in hte tube will compress the air inside the tube and will produce extra pressure `DeltaP(say)`. The length of air column in the tube will be =`(1-h)`. if P is the pressure of the compressed air in the tube, then using Boyal's law, we have `P(l-h) =P_(0)l` or `1P= P_(0)l(l-h)` `:. Delta P = P-P_(0)=(P_(0)l)/(l-h) P_(0)=(P_(0)h)/(l-h)` The EXCESS pressure `(=2 S//r)` produced by surface tension should be equal to the sum of pressure created by water column of height h and air compressed in the capillary i.e, `(2S)/(r) = Delta P + h rho G = (P_(0)h)/(l-h) + h rho g` On solving, we get `l = (P_(0)rh)/(2S-h rho g r)+h` `=((1.01xx10^(5))xx(0.05 xx 10^(2)) xx (1 xx 10^(-2)))/(2 xx 970 xx 10^(-3)) - (1 xx 10^(-2)) xx 10^(3) xx 9.8 +0.01` `xx(0.05 xx 10^(-2))` `=5.52m`. |
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| 14733. |
A constant horizontal force of 20 N acts on a body on a smooth horizontal plane. The body starts from rest and is observed to move 20 m in two seconds. The mass of the body is |
| Answer» ANSWER :B | |
| 14734. |
A metallic wire of diameter “d” is lying horizontally on the surface of water. The maximum length of wire so that is may not sink will be |
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Answer» `SQRT((2T)/( pi d g))` |
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| 14735. |
Imagine a hard surface along xz plane to be fixed. A particle moving along the line 4x + 3y - 12 = 0 with a speed of 10m/s from the positive side of y-axis, approaches towards the plane and collides. If the coefficients of restitution be e = 0.75, then the speed of the particle after collision will besqrt(2)xm//s, then x = ......... |
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| 14736. |
Three moles of an ideal gas at a pressure P_(A) and temperature T_(A)is isothermally expanded to twice its initial volume. It is then compressed at constant pressure to its original , volume. Finally the gas is compressed at constant volume to its original pressure P_(A) (a) Sketch the P-V and P-T diagrams for the complete process. (b) Calculate the net work done by the gas, and net heat supplied to the gas during the complete process. |
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Answer» Solution :The P-V and P-T diagrams are as shonw in (a) and (b) A is the initial STATE. The FIRST second and third processes are represented by AB, BC and CA respectively. (b) WORK done during the isothermal process `AB= nRT log _(e) (V_(2)//V_(1))` `= 3 xx 8.3 xx T xx 2. 303 xx log _(10) (2V//V) = 17.26 TJ` Work done during isobaric process `BC = (P//2) (V _ 2V) =- PV //2` `=- 3 xx 8.3T//2 =- 12 . 45 TJ.`  Work done during the isochoric process `CA =0` because `dV =0.` Net work done `=17. 26T - 12. 45 T + 0 = 4.81 TJ` The change in internal energy `= Delta U = 0` because the final state is the same as initial state. By first LAW `Delta Q = Delta U + DeltaW=0 + 4.81 T = 4.81 TJ` |
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| 14737. |
A metal ball suspended from a spring balance is immersed in water at 4^(@) C. If the temperature of water is changed the reading in the balance (a) may decreases (b) increases (c) may remains same |
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Answer» only (C) is TRUE |
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| 14738. |
Why blood pressure in human is greater at the feel than at the brain? |
| Answer» Solution :Pressure `P=hrhog`. The height h of the blood column at the FEET is LARGER COMPARED to that at the BRAIN. So blood pressure is more at the feet than at the brain. | |
| 14739. |
A spherical liquid drop of radius R is divided into eight equal droplets. If surface tension is T, the work done in the process will be |
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Answer» `2piR^2S` |
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| 14740. |
A particle is dropped down in a deep hole which extends to the centre of the earth. Calculate the velocity at a depth of one km from the surface of this carth ? Assume that g= 10 m//s^2and radius of the earth = 6400 km. |
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Answer» Velocity `v=omegasqrt(a^2-y^2)=0.00215sqrt((6.4xx10^6)-(6.399xx10^6)^2)=1.44xx10^(12)MS^(-1)` |
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| 14741. |
Friction arises on account of……..at the …….. |
| Answer» Solution :adhesive pressure , POINTS of ACTUAL CONTACT . | |
| 14742. |
A vehicle of mass 120kg is moving with a velocity of 90 kmph . What force should be applied on the vehicle to stop it in 5s. |
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| 14743. |
(A) : A measurement which is less accurate is more precise. (R) : Precision tells only about the resolution of measurement but not about accuracy. |
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Answer» Both (A) and (R) are true and (R) is the CORRECT explanation of (A) |
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| 14744. |
Three samples of the same gas x, y and z, for which the ratio of specific heats is r = 3/2, haveinitially the same volume. The volumes of each sample is doubled, by isobaric process in the case of y and by isothermal process in the case of z. If the initial pressures of the samples x,y and z are in the ratio 2sqrt2:1:2, then the ratio of their final pressures is |
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Answer» `2:1:1 ` |
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| 14745. |
Two identical cars A and B are moving at 36 kmph. A goes on a bridge, convex upward and .B. on concave upward. If the radius of curvature of bridge is 20 m, the ratio of normal forcesexerted on the cars whenthey are at the middle of bridges (g = 10 ms^(-2)) |
| Answer» ANSWER :A | |
| 14746. |
A black body is kept in a hollow enclosure maintained at same temperature as that of the black body. If the black body is replaced by a non-black body of same shape and size, maintained at same temperature, then total radiation emitted by the non-black body |
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Answer» is same as that of the black BODY |
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| 14747. |
AB is a quarter of a smooth circular track of radius 4 m as shown in the figure. A particle P of mass 5 kg moves along the track from A to B under the action of the following forces (1) A force F_(1) directed always towards point B, its magnitude is constant and equals 4 N, (2) A force F_(2) that is directed along the instantaneous tangent to the circular track, its magnitude is (20-s) newton where s is the distance travelled in metre. (3) A horizontal force F of magnitude 25 N. Find the workdone by each force. |
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Answer» Solution :(i) Work done by FORCE `F_(1)`As shown in figure, let the particle be at point P at some instant of time t = t. The particle moves from position P to position Q in small interval of time dt. The direction of force on particle P will be in the direction PB. The small amount of work done in time dt is `dW_(1) = F_(1).ds = F_(1) COS theta.ds = F_(1) cos theta.Rd theta` `because ds = Rd theta = 16 cos theta d theta ""(because R = 4m and F = 4N)` Now the total work done as the particle moves from A and B is given by `W_(1) = int_(0)^(PI//4) 16 cos theta d theta = 16 [SIN theta]_(0)^(pi//4) = 16 ((1)/(sqrt(2))) = (8 sqrt(2)) = 11.30` joule .......(1) (ii) Work done by force `F_(2)` In the case, `d W_(2) = F_(2) xx ds = (20 - s)ds` But `s = R theta and d s = Rd theta ""therefore d W_(2) = (20 - R theta) = (20 - 4 theta) xx 4 d theta` `W_(2) = int_(0)^(pi//4) (80 d theta - 160 d theta) = 80 [theta]_(0)^(pi//4) - 16 [(theta^(2))/(2)]_(0)^(pi//4) = (80 xx pi)/(4) - (16)/(2) ((pi)/(4))^(2) = (62.83 - 4.93) = 57.9 J` ....... (2) (iii) Work done by force `F_(3)` The magnitude of `F_(3)` is 25 N which is always horizontal. The net displacement of the particle is OB. Hence the work done `W_(3) = 25 xx 4 = 100` joule ...........(3) |
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| 14748. |
When a motor cyclist takes a U-turn in 4s what is the average velocity of the motor cyclist? |
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Answer» Solution :When the MOTOR cyclist TAKES a U-turn, angular displacement, `theta = pi rad and t = 4 s `. The AVERAGE angular velocity, `omega = (theta)/(t)=(pi)/(4) = 7.855xx10^(-1) = 0.7855rads^(-1)`. |
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| 14749. |
The perssure exerted on the walls of the container by a gas is due to the fact that the gas molecules ............... . |
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Answer» LOSE there K.E |
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| 14750. |
Four identical particles each of mass 1 kg are arranged at the corners of a square of side length 22 m. If one of the particles is removed, the shift in the centre of mass is |
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Answer» `(8)/(3)m` |
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