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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.
| 14851. |
If the presure of an ideal gas contains in a closed vessel is increased by 0.5%, the increase in temperature is 2^(@)C. The initial temperature of the gas is |
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Answer» `27^(@)C ` |
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| 14852. |
Define angle of contact for a given pair of solid and liquid. |
| Answer» SOLUTION :The ANGLE between TANGENTS drawn at the point of contact to the liquid surface and solid surface inside the liquid is call the angle of contact for a PAIR of solid and liquid for stop it is denoted by `theta`. | |
| 14853. |
Find the value of 31.2 -12.125with due consideration of significant figures. |
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Answer» SOLUTION :`31.2-12.125` Here `31.2` has ONE DECIMAL place, HENCE `12.125`is to be rounded off to `1+1=2`decimal places `12.125`is rounded off as `12.12` `:. 31.2-12.12=19.08` finally we should have only one decimal place and hence `19.08`ISTO be rounded off as `19.1`. |
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| 14854. |
A body of mass m is suspended from a weightless wire of length l. If Y is the Young's modulus of the material of the wire, calculate the frequnecy of vibration in the vertical direction. |
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Answer» Solution :If the mass m causes an elongation x of the WIRE, then LONGITUDINAL stress = Young.s modulus `xx` longitudinal strain = `Y*x/l`. By definition, longitudinal stress = REACTION force acting per UNIT area `therefore` Reaction force = longitudinal stress `xxalpha" "[alpha="cross-sectional area of the wire"]` = `(Yx)/lxxalpha=(Yxalpha)/l`. This reaction force acts as the restoring force. So, ACCELERATION of mass m, `a=("restoring force")/m=(Yxalpha)/(ml)` `therefore` Frequency, `n=1/(2pi)sqrt(("acceleration")/("displacement"))=1/(2pi)sqrt(alpha/x)=1/(2pi)sqrt((Yalpha)/(ml))`. |
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| 14855. |
In a uniform electric field, find the total flux associated with the fiven surface. |
| Answer» SOLUTION :ZERO in all the CASES | |
| 14856. |
Figure 6.19 shows the p-T graphs for a fixed mass of an ideal gas at volumes V_1 and V_2.Can it be concluded from the graphs that V_1 is greater than V_2? |
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Answer» Solution :`pV=nRT=m/M RT` or, `p=(mR)/(MV)T`[ M=molecular WEIGHTS] So p-T GRAPHS are STRAIGHT lines PASSING through the origin,for any fixed V . The slope `(mR)/(MV)` is smaller for a greater value of V. In fig.6.19 the slope for `V_1` is smaller than that for `V_2`. So, `V_1 gt V_2` 
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| 14857. |
Define power, average power, instantaneous power. |
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Answer» Solution :Power: Power is defined as the RATE of work done or energy delivered. Power (P) = `("Work done (W)")/("TIME taken (t)")` `P=(W)/(t)` Average power: The average power `(P_(av))` is defined as the ratio of the total work done to the total time taken. `P_(av)=("Total work done")/("Total time taken")` INSTANTANEOUS power: The instantaneous power `(P_("inst"))` is defined as the power delivered at an instant (as time interval APPROACHES zero). `P_("inst")=(dW)/(DT)` |
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| 14858. |
A constant retarding force of 50 N is applied to a body of mass 20 kg moving initially with a speed of 15 ms^(-1) . How long does the body take to stop? |
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Answer» Solution :Acceleration `a = (-F)/(m) = (-50)/(20) = - 2.5 ms^(-2)` ` u= 15 ms^(-1), V= 0 , t = ? ` ` t =(15)/(2.5) = 6s` |
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| 14859. |
When a motor cyclist takes a U-turn in 4s what is the average angular 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 = 4s. The AVERAGE angular velocity, `omega = (theta)/(t) = (pi)/(4) = 7.855 xx 10^(-1) = 0.7855 " rad s"^(-1)`. |
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| 14860. |
A stone is thrown vertically upward with an initial velocity of 10ms^(-1)from the top of a cliff of height40m. How much time it take to reach the ground? |
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| 14861. |
A pendulum based clock keeps correct time in an aeroplane flying uniformly at a height h above the surface of the earth. The cabin temperature inside the plane is 10^(@)C. The same pendulum keeps correct time on the surface of the earth when temperature is 30^(@)C. Find the coefficient of linear expansions of the material of the pendulum. You can assume that h ltlt R (radius of the earth) |
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| 14862. |
Speed of sound wave in a fluid depends upon……….. |
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Answer» directly on density of the medium. (where B = Bulk modulus, `rho=` density of medium) `implies v prop sqrtB implies` Option (D) is TRUE. `THEREFORE v prop (1)/(sqrt rho)implies` Option © is true. |
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| 14863. |
The time period of a satellite of Earth is 5 hours. If the separation between the Earth and the satellite is increased to four times the previous value, the new time period will become. |
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Answer» Solution :`T_(2)=T_(1)((a_(2))/(a_(1)))^(3/2) = T_(1)[(4a_(1))/(a_(1))] ^(3/2) = 5(4)^(3/2) = 5(8)` `T_(2) = 40` hours |
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| 14864. |
Two ships A and B are 10km apart on a line running south to north. Ship A farther north is streaming west at 20 km/h and ship B is streaming north at 20 km/h. Their distance of closest approach and how long do they take to reach it ? |
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Answer» `5sqrt(2)`KM, 15 MIN |
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| 14865. |
What is meant by " tensile stress " ? |
| Answer» SOLUTION :The MAXIMUM STRESS ulitimate stress point beyond which the wire breaks is called breaking stress or tensile strength. | |
| 14866. |
Velocity of a particle is given by v = (3t ^(2) +2t) m/s. Find its average velocity between t = 0 to t=3s and also find its acceleration at t = 3s. Motion of the particle is in one dimension. |
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Answer» `11 m /s, 10 (m)/(s ^(2))` `therefore int v dx = int (3t ^(2) + 2r ) dt` `therefore x = 3 (t ^(3))/(3) + 2 (t ^(2))/( 2) = t ^(3) + t ^(2)` At `t =0` TIME ,` x _(0) = (0) ^(2) + (0) ^(2) =0` At t =3 time, `x _(3) = (3) ^(2) + (3) ^(2) = 36 m` `therefore` Average velocity `= (x _(3) -x _(0))/(3-0)=(36-0)/(3-0)` `=12 MS ^(-1)` Acceleration of PARTICLE, `a = (dv)/(dt) = (d)/(dt) (3t ^(2) + 2t) = 6T +2` Put `t =3s,` `a=6(3) +2=20 ms ^(-2)` |
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| 14867. |
If the absolute temperature of a gas is increased 3 times the rms velocity of the molecules will be ............. . |
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Answer» 3 TIMES `V_(rms1) = sqrt(KT) , V_(rms2) = sqrt(3KT)` `(V_(rms2))/(V_(rms1)) = (sqrt(3KT))/(sqrt(KT)) = sqrt3` `V_(rms2) = sqrt3 V_(rms1)` |
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| 14868. |
At what temperature will the oxygen molecules have the same root mean square speed as hydrogen molecules at 200K |
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Answer» 800K |
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| 14869. |
When the maximum energy during elastic collision will be exchanged ? |
| Answer» SOLUTION :Maximum energy will be exchanged when two bodies of EQUAL MASS COLLIDES . | |
| 14870. |
A block can slide down an a smooth inclined plane of inclination theta, kept on a floor of a lift. When the lift is descending with a retardation a. The acceleration of the block relative to the incline is |
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Answer» `a-g SIN THETA` |
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| 14871. |
A machine delivers power given by P=(p_(0)t_(0)^(2))/((t+t_(0))^(2)) where p_(0) are constants. The machine starts at t=0 and runs forever. What is maximum work that the machine can perform. |
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Answer» <P>Infine |
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| 14872. |
Value of acceleration due to gravity is 9.8m//sec^(2). Find its value in km//hr^(2). |
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Answer» Solution :`n_(1)=n_(2)[(L_(2))/(L_(1))]^(1)[(T_(2))/(T_(1))]^(-2)`. Given `n_(2)=9.8` `therefore n_(1)=9.8 [(1M)/(1km)]^(1)[(1 SEC)/(1HR)]^(-2)` `=9.8[(1m)/(1000m)]^(1)[(1 sec)/(60xx60sec)]^(-2)` `n_(1)=9.8[(1)/(1000)xx60xx60xx60xx60]` `n_(1)=98xx36xx36=127008` `therefore g=127008km//hr^(2)` |
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| 14873. |
A particle revolves in a circle of radius 0.1 m, making 300 rpm. Calculate its linear velocity and centripetal acceleration. |
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| 14874. |
A chain of length 'L' and mass 'M' is hanging by fixing its upper end to a rigid support. The tension in the chain at a distance 'x' from the rigid support is |
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Answer» Zero |
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| 14875. |
Show that the area of the triangle contained between the vectors veca and vecb is one half of the magnitude of vecaxxvecb. |
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Answer» Solution :In `DeltaOAB, vec(OA)=VECB, vec(OB)=veca` and `angleAOB=theta and AC=h` `:.` from `DeltaACO` `(AC)/(OA)=SINTHETA` `:. H=bsintheta` Now AREA of `DeltaOAB=(1)/(2)xx|vec(OB)|h` `=(1)/(2)|veca|h` `:.` Area = `(1)/(2)absintheta [because h=bsintheta]` `=(1)/(2)|veca||vecb|sinthetahatn` `:.vecS=(1)/(2)(vecaxxvecb)hatn` Magnitude `S=(1)/(2)|vecaxxvecb|` 
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| 14876. |
The specific heat of Argon at constant volumeis 0.3122kJ/kg K. Find the specific heat of Argon at constant pressure if R = 8.314 kJ/k mole K. (Molecular weight of argon = 39.95) in kJ/kg K is. |
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Answer» 520.3 |
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| 14877. |
If 'm' is mass 'Q' is charge and B is magnetic induction, m/BQ has the same dimensions as |
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Answer» FREQUENCY |
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| 14878. |
A particle traversed half of the distance with a velocity of V_(0).The remaining parts of the distance was covered with velocity V_(1), for half of the time and with V_(2) for other half of the time .Find the mean velocity of the particle averahed and the whole time of motion . |
Answer» Solution : Average VELOCITY for the second half distance `=(v_(1)""t/2+v_(2)""t/2)/(t/2+t/2)=(v_(1)+v_(2))/(2)` Average velocity for the first half distance `=v_(0)` Average velocity for total PATH `=(2v_(0) ((v_(1)+v_(2))/(2)))/(v_(0)+(v_(1)+v_(2))/(2))= (2v_(0) (v_(1)+v_(2)))/(v_(1)+v_(2)+2v_(0))` |
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| 14879. |
What is meant by rolling friction ? |
Answer» Solution : The invention of the wheel plays a CRUCIAL role in Rolling friction human civilization. One of the important applications is suitcases with rolling on coasters. Rolling wheels makes it easier than carrying luggage. When an object moves on a surface, ESSENTIALLY it is sliding on it. But wheels move on the surface through rolling motion. In rolling motion when a wheel moves on a surface, the Rolling and kinetic friction point of contact with surface is ALWAYS at rest. Since the point of contact is at rest, there is no relative motion between the wheel and surface. HENCE the frictional force is very less. At the same time if an object moves without a wheel, there is a relative motion between the object and the surface. As a result frictional force is larger. This makes it difficult to move the object. The figure shows the difference between rolling and kinetic friction. IDEALLY in pure rolling, motion of the point of contact with the surface should be at rest, but in practice it is not so. Due to the elastic nature of the surface at the point of contact there will be some deformation on the object at this point on the wheel or surface as shown in figure. Due to this deformation, there will be minimal friction between wheel and surface. It is called rolling friction . In fact, rolling frictionis much smaller than kinetic friction. |
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| 14880. |
P is a point at a distance r from the centre of a spherical shell of mass M and radius a, where r lt a. The gravitational potential at P is |
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Answer» `-(GM)/(R)` |
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| 14881. |
A particle P is projected Vertically upward from a point A . Six seconds later, another particle Q is projected vertically upward from A . Both P and Q reach A simultaneously. The ratio o f maximum heights reached by P and Q=65 : 25. Find thevelocity of the projection of Q in m//s. |
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Answer» SOLUTION :`1/2 G(t+3)^(2), 1/2 "GT"^(2) =64:25 `(or) `(t+3)^(2):t^(2)=64:25,`(or)`(t+3):t=8:5` `5t+15=8t`(or)`3t=15,t=5sec` `v=gxxt = 9.8xx5=49m//s=5g m//s` |
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| 14882. |
The s-t graph of a particle moving with constant acceleration at time t, makes an angle 45^(@) with the time axis. After 1 second the angle changes to 60^(@). Find the acceleration of the particle. |
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| 14883. |
A physical quantity, U can be measured as U=(abz^2)/(h^3). Write the expression for finding the relative error in U. |
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Answer» Solution :RELATIVE ERROR in U can be measured as : `(TRIANGLEU)/(U)=(trianglea)/(a)+(triangleb)/(b)+2(trianglez)/(Z)+3(triangleh)/(h)`. |
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| 14884. |
A vertical disc has three grooves directed along chords AB, AC and AD. Three bodies begin to slide down the respective grooves from A simultaneously. If AB gt AC gt AD, the respective time intervals to reach the bottom of the respective groovest_(1), t_(2) and t_(3) are |
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Answer» `t_(1) GT t_(2) gt t_(3)` |
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| 14885. |
The distance between two moving particles at any time is a. If v be their relative velocityand v_(1) andv_(2) be the components of v along and perpendicular to a. The time when they are closest to each other is |
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Answer» `( a v_(1))/(V^(2))` |
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| 14886. |
Three balls A,B,Cof masses m_1 ,m_2 and m_3 respectively are kept at rest along a straightline. Now A moving in that straight line with velocity u_1strikes C. As a rsult velocity of C becomes u_3. If the collisions areelastic , show that u_3 = 4u_1when m_1gt gt m_2 and m_2gt gt m_3. In case A hits C directly will the velocity of C be higher or lower ? |
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Answer» Solution :Suppose A acquires a velocity `v_1` after collision with B. From the law of conservation of linear momentum , for collision between A and B . `m_1u_1=m_1v_1+m_2u_2` `or, m_1(u_1-v_1)=m_2u_2` From the law of conservation of kinetic energy of elastic collision, `1/2 m_1u_1^2=1/2m_1v_1^2+1/2m_2u_2^2` `or, m_1(u_1^2-v_1^2)=m_2u_2^2` DIVIDING equation (2) by (1) , `u_1+v_1=u_2 or, v_1=u_2-u_1` Inserting the value of `v_1` in equation (1) , `m_1(2u_1-u_2)=m_2u_2 or, 2m_1u_1=(m_1+m_2)u_2` Hence, `u_2=(2m_1)/(m_1+m_2)*u_1=2/(1+(m_2)/m_1)*u_1` As `m_1 gt gt m_2 , (m_2)/(m_1)lt lt 1,hence, 1+(m_2)/(m_1)~~1`. THUS , `u_2= 2u_1`. Similary for the collisionof B iwith C , `u_3 ~~2u_2`. thus `u_3= 4u_1`. But , for the direct collision of A with C ,the velocity GAINED by C, `u_3^.~~2u_1`. `therefore u_3^.` isless than `u_3`. Hence, in case of direct collision between A and Cthe speed of C will be less. |
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| 14887. |
A crown glass of refracting angle 8^@ is combined with a flint glass prism to obtain deviation without dispersion. If the refractive indices for red and violet rays for crown glass are 1.514 and 1.524 and for the flint glass are 1.645 and 1.665 respectively, Find the angle of flint glass prism and net deivation. |
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Answer» Solution :The CONDITION for deviation with out DISPERSIONIS `(mu_v-mu_R) A=(mu_v-mu_R)A.` `therefore A.=((1.524 -1.514) times 8^@)/((1.665 -1.645))=0.08^@/0.02=4^@` for crown glass `MU=(1.514+1.524)/2=1.519` for FLINT glass `mu=(1.645+1.665)/2=1.655` `therefore ` The net deviation `(delta-delta.)=(mu-1)A-(mu.-1)A.` `=0.159 times 8^@-0.655 times 4^@=1.53^@` |
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| 14888. |
If the earth did not have an atmosphere, it would become intoleraby cold. Why? |
| Answer» Solution :Due to GREEN HOUSE effect, the presence of atmosphere prevents HEAT radiations RECEIVED by earth to go back. In the ABSENCE of atmosphere radiation will go back at night making the temperature very low and inhospitable. | |
| 14889. |
Find the displacement of the wedge when m comes out of the wedge. There is no friction anywhere. |
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Answer» SOLUTION :Let wedge move TOWARDS RIGHT by DISTANCE `x` then `/_\x_(CM)=(m(lcostheta+x)+Mx_m(x-l))/(m+M+m)` `0=mlcostheta-ml(m+M+m)x` `x=(ml-mlcostheta)/(2m+M)` |
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| 14890. |
Two equal spheres A and B lie on a smooth horizontal circular groove at opposite ends of a diameter. At time t = 0, A is projected along the groove and it first impinges on B at time t = T_(1) and again at time t = T_(2). If .e. is the coefficient of restitution, find the ratio of T_(2)//T_(1) |
Answer» Solution : `T_(1)=(PIR)/(u_1)"" ....(1)` `(v_(2)-v_(1))/(u_(1))=erArrv_(2)-v_(1)=eu_(1)` Time taken to COLLIDE A and B again, `T_(2)-T_(1)=(2piR)/(v_(2)-v_(1))rArrT_(2)-T_(1)=(2piR)/(eu_(1)) "" ....(2)` `(2)DIV(1)RARR(T_(2))/(T_(1))=(2+e)/(e)` |
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| 14891. |
A particle of mass m_(1) is fastened to one end of a massless string and another particle of mass m_(2) is fastened to the middle point of the same string. The other end of the string being fastened to a fixed point on a smooth horizontal table. The particles are then projected, so that the two particles and the string are always in the same straight line and describe horizontal circles. Then, the ratio of rotations in the two parts of the string is : |
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Answer» `m_(1)//(m_(1)+m_(2))` |
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| 14892. |
A wooden block of volume 1000 cm^(3) is suspended from a spring balance. It weighs 12 N in air. It is suspended in water such that half of the block is below the surface of water. The reading of the spring balance is |
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Answer» 10 N |
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| 14893. |
Give two points of similarity between viscous force and solid friction? |
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Answer» SOLUTION :(i) Both come into PLAY WHENEVER there is relative motion. (ii) Both OPPOSES the motion. |
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| 14894. |
Which of the following statements are true for a stationary wave ? |
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Answer» (A) EVERY particle has a fixed AMPLITUDE which is different from the amplitude of its nearest particle. For A given particle x = constant A.= constat For neightbouring particle, x is SLIGHTLY different . Hence , their amplitudes are also slightly different. Thus, option (A) is true. (b) Here frequency f is same for all the particles. Hence, T is same for all the particles and so `T/2` is also same which is time to pass through the respective mid point. Thus, option (B) is true. (c) All the particles do not have same ampli tudes because here amplitude of any given particle depends on its position according to formula A. = 2a sin(kx). Thus, option © is false. (d) In case of stationary wave, there is no trans fer of energy because the entire length of medium becomes an isolated system which does not interact with the surrounding. Thus, option (D) is true. (e) At all the nodal points, particles of medium remain at rest at all the times. Thus, option (E) is true. |
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| 14895. |
A circular ring of radius 7 cm is supported horizontally from the pan of a balance, so that it comes in contact with the water in a glass vessel. What force will required to detach it from the surface of water. (Surface tention of water =0.072N/m). |
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Answer» Solution :Neglecting the thickness of the ring, the net DOWNWARD force on the INNER and OUTER circumferenc of the ring dur to SURFACE tesion = `Sxx2pi(R+R)=Sxx4piR` `S=0.072N//m, R=7cm=7x10^(-2)m` The vertical force required to detach it from the water surface is `F=Sxx4piR` `=0.072xx4xx22/7=10^(-2)=6.336xx10^(-2)N` |
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| 14896. |
A straight piece of wire 5 cm long is placed horizontally on the surface of water and gently pulled up. It is found that a force of 7 xx 10^(-3) N in addition to weight of wire is required for that purpose. The surface tension of water is |
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Answer» `7 xx 10^(-2)N//m` |
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| 14897. |
A man starts walking from a point on the surface of earth (assumed smooth) and reaches diagonally opposite point. What is the nature of work done by him? |
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Answer» zero SINCE `d=(-x+x)=0` (DIAGONALLY OPPOSITE POINT) |
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| 14898. |
A barometer with a brass scale correct at 0^(@)C reads 70 cm of mercury on a day when the air temperature is 40^(@)C. The correct reading at 0^(@)C is (Coefficient of real expansion of mercury is 0.00018//^(@)C and the coefficient of linear expansion of brass is 0.000018//^(@)C. |
| Answer» SOLUTION :`69.55 CM` | |
| 14899. |
The dimensions of gravitational constant are |
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Answer» `[M^-1L^3T^-2]` |
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| 14900. |
Science is ……. Which humans have gained through ………….. |
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Answer» |
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