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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.
| 14401. |
Identify the correct order in which the values of M.I. decreases for the following (i) M.I. of solid sphere of mass .M. and radius "R. about its diameter of rotation (ii) M.I of uniform ring of mass .M. and radius .R. about its tangent perpendicular to its plane(iii) M.I. of uniform dise of mass .M. and radius .R. about its diameter (iv) M.I of a uniform solid cylinder of mass M about its own axis of rotation |
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Answer» III,I,iv,II |
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| 14402. |
If vec(A)+vec(B)=vec(C ), then magnitude of vec(B) is |
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Answer» `VEC(C )-vec(A)` |
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| 14403. |
A spherical steel ball released at the top of a long column of glycerine of length l falls through a distance l//2 with accelerated motion and the remaining distance l//2 with uniform velocity V. Let t_(1)andt_(2) denote the times taken to cover the first and second half and W_(1)andW_(2) the work done against gravity in the two halves, then compare times and work done. |
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Answer» Solution :The AVERAGE velocity in the first half of the distance `ltv`, while in the second half the average velocity is v. Therefore, `t_(1)gt t_(2)`. The work done against gravity in both HALVES is `mg l//2` `therefore t_(1)gtt_(2)w_(1)=w_(2)` |
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| 14404. |
In a simple harmonic oscillation, the acceleration against displacement for one complete oscillation will be |
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Answer» an ELLIPSE |
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| 14405. |
The time period of simple pendulum is 'T'. When the length increases by 10 cm, its period is T_(1). When the length is decreased by 10 cm, its period is T_(2). Then the relation between T, T_(2) and T_(2) is |
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Answer» `(2)/(T^(2))= (1)/(T_(1)^(2))+ (1)/(T_(2)^(2))` |
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| 14406. |
Can the position-time graph have a negative slope? |
| Answer» SOLUTION :YES when the VELOCITY of the OBJECT is NEGATIVE . | |
| 14407. |
A vector vec(L)(t) of fixed magnitude lies in XY-plane. It rotates at a constant angular velocity of vec(w)=(3"rad/s")hat(k). Find the value of (d)/(dt)(vec(L)(t)) at the instant when vec(L)=3hat(i)+4hat(j) |
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Answer» `-12hat(i)+9ahat(J)` |
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| 14409. |
A flat spiral spring is stretched by mens of a small weight . The spring undergoes |
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Answer» longitudinal STRAIN and shear strain |
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| 14410. |
In a compound micrscope, if the objective produces an image I_0 and the eye piece produces an image I_e, then |
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Answer» `I_0` is VIRTUAL but `I_e` is REAL |
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| 14411. |
Maximum velocity and maximum acceleration of a particle executing SHM are beta" and "alpha respectively, what will its frquency? |
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Answer» Solution :`implies (a_("MAX"))/(v_("max"))= (A OMEGA^(2))/(A omega)` `(alpha)/(BETA)= omega` `therefore (alpha)/(beta)= 2pi F` `therefore f= (alpha)/(2pi beta)`. |
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| 14412. |
Two rods A and B are of equal lengths. Their ends are kept at the same temperature difference and their area of cross - sections are A_(1) and A_(2) and thermal conductivities k_(1) and k_(2). The rate of heat transmission in two rods will be equal, if. . . .. . . |
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Answer» `k_(1)A_(2)=k_(2)A_(1)` `:.k_(1)A_(1)((T_(1)-T_(2))/(L))=k_(2)A_(2)((T_(1)-T_(2))/(L))` `:.k_(1)A_(1)=k_(2)A_(2)` |
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| 14413. |
A wire of length 'L' and cross sectional area 'A' is made up of a material of young's modulus 'Y'. If the wire is stretched by 'X' the work done is |
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Answer» `(YAX)/(L)` |
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| 14414. |
Statement A : If the force varies with time in a complicated way then the average force is measured by the total change in momentum of the bodyStatement B : Change in momentum and inpulsive force are numerically equal |
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Answer» A & B are true |
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| 14415. |
A ship moves due to east at 12km/hr for one hour and then turns towards exactly towards south to move for an hour at 5km/hr. Calculate its magnitude of average velocity for the given motion. |
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Answer» |
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| 14416. |
A wooden block with a coin placed on its top floats in water with 1 as the length immersed and h as the height of water column. After some time the coin falls into water. Then |
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Answer» 1 DECREASES and H INCREASES |
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| 14417. |
A car weighing 500 kg climbs up a hill of slope 1 in 49 with a velocity of 36 KMPH. If the frictional force is 50 N, the power delivered by the engine is |
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Answer» 500W |
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| 14418. |
Match the following given in both columns correctly .{:("Column I","Column II"),(1."Centripetal force","(i)radius" xx"angular velocity"),(2."Centrifugal force","(ii)"1//2 at^(2)),(3."Linear velocity v ","(iii)mass" xx "velocity"),(4."Momentum","(iv)mass" xx "acceleration"),(,"(v)towards the centre of the circle"),(,"(vi)away from the centre of the circle"):} |
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Answer» 1 - (v), 2 - (i), 3 - (iv), 4 - (iii) |
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| 14419. |
Which of the following functions of time represent (a) simple harmonic motion and (b) periodic but not simple harmonic motion? Give the period for each case. (i) sinomegat-cosomegat (ii) sin^(2)omegat (iii) cosomegat+2sin^(2)omegat |
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Answer» a simple HARMONIC MOTION with a period `(pi)/(omega)`. `=sqrt(2)(sinomegat"cos"(pi)/(4)-cosomegat"sin"(pi)/(4))=sqrt(2)sin(omegat-(pi)/(4))` It represented SHM with a period `(2pi)/(omega)`. |
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| 14420. |
For a person near point of vision is 1010cm. Then the power of lens he must wear so as to have normal vision, should be |
| Answer» ANSWER :C | |
| 14421. |
A band playing music at frequency f is moving towards a wall at a speed ofv_(b) . A motorist is following the band with a speed ofv_(m). If v is the velocity of sound , obtain an expression for the beat frequency heard by the motorist |
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Answer» |
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| 14422. |
A thread is tightly wrapped on two pulleys as shown in figure. Both the pulleys are uniform disc with upper one having mass M and radius R being free to rotate about its central horizontal axis. The lower pulley has mass m and radius r and it is released from rest. It spins and falls down. At the instant of release a small mark (A) was at the top point of the lower pulley. (a) After what minimum time (t_(0)) the mark will again be at the top of the lower pulley? (b) Find acceleration of the mark at time t_(0). (c) Is there any difference in magnitude of acceleration of the mark and that of a point located on the circumference at diametrically opposite end of the pulley. |
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Answer» (b) `a = (2g)/(2m + 3M)N sqrt(m^(2) + (4 pi M + M + m)^(2))` (c) yes |
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| 14423. |
Calculate the energy radiated per minute by ablack body of surface area 200 cm^(2) , maintained at 127^(@)C. sigma = 5.7xx10^(-8)Wm^(-2)K^(-4) |
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Answer» |
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| 14424. |
A Carnot engine is being operated by taking heat from a source at temperature 527^(@)C. If the surrounding temperature is 27^(@)C, what is the efficiency of the engine? If the source supplies heat at the rate of 10^(9)J per minute, how much usable work is obtained per minute? |
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Answer» Solution :Temperature of the source, `T_(1) = 527^(@)C = (527 + 273) K = 800 K` Temperature of the sink, `T_(2) = 27^(@)C = (27+ 273) K = 300 K` So, efficiency `ETA = W/(Q_1)` `:. W = eta Q_(1) = 5/8 xx 10^(9) = 6.25 xx 10^(8) J cdot "min"^(-1)`. |
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| 14425. |
for a satellite to be geostationary, which of the following are essential conditions? (i) It must always be stationed above the equator. (ii) It must rotates from west to east. (iii) It must be about 36,000 km above the earth (iv) Its orbit must be circular and not elliptical. |
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Answer» (i),(II) |
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| 14426. |
The trajectory of a projectile in a vertical plane is y=ax-bx^(2), where a and b are constants and x and y are respectively horizontal and vertical distance of the projectile from the point of projection. The maximum height attained by the particle and the angle of projection from the horizontal are |
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Answer» Solution :`y=ax-bx^(2)` For height (or y) to be MAXIMUM `(dy)/(dx)=0` or `a-2bx=0` (or) `x=a/(2b)` `:.y_("max")=a(a/(2b))-B(a/(2b))^(2)=(a^(2))/(4b)`…..ii `((dy)/(dx))_(x=0)=a=tan theta_(0)""` where `theta_(0)` is the angle of projection `:.theta_(0)=tan^(-1)` a. |
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| 14427. |
Two identical point masses are placed at separation of d. Out of the following graphs the graph that represent the variation of gravitational field intensity E with the distance r from any one mass to the other along the line joining them is |
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Answer»
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| 14428. |
A binary star consists of two stars A (mass 2.2M_(s)) and B (mass 11m_(s)) where M_(s) is the mass of the sun. They are separted by distance d and are rotating about their centre of mass, which is stationary. The ratio of the total angular momentum of the binary star to the angular momentum of star B about the centre of mass is. |
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| 14429. |
The absolute temperature of a body A is four times that of another body B. For the two bodies, the lifference in wavelengths, at which energy radiated is maximum is 3.0mu m. Then the wavelength at which the body B radiates maximum energy, m micrometers is |
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Answer» 2 |
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| 14430. |
Find the velocity of centre of mass of the system of two moving particles of mass m and 2m, as shown in the figures (i), (ii) and (iii). |
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Answer» Solution :Referring to the equation `vecv_(cm) = (summ_(1)vecv_(1))/(summ_(1))` we have, (i) `vecv_(cm) =((m)(2vhati) + (2m)(-VHATI))/(m+2m) =VEC0` (ii) `vecv_(cm) =(m(2vhati ) + 2m(vhati))/(m+2m) =(4v)/3 hati` (iii) `vecv_(cm) = (m(2vhati ) + 2m(vhatj))/(m+2m) =(2V)/3(hati + hatj)` |
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| 14431. |
A weighted glass tube is floated in a liquid with 20 cm of its length immersed. It is pushed down through a certain distance and then released. Compute the time period of its vibration. |
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Answer» Solution :Height to which the tube is immersed in the liquid = h=20 cm Let the mass of the tube bem and the density of the liquid bep. The tube floats when the weight of the tube equals the upthrust (see FIG.) `mg = 20 xx A xxrhog` `m = 20 xx Arho`where A is the AREA of cross-section of the tube. Let y be the depth through which the tube is pushed into the liquid. Restoring force on the tube =`yArhog` Acceleration of the tube `=( yArhog)/(20 Arho)` = `(YG)/20`  The acceleration is proportional to the displacement. So the motion of the tube is shm. The period of OSCILLATION is given by `T=2pisqrt(("Displacement")/("Acceleration"))` `= 2pisqrt(y/(yg//20))` `y = 20 cm0.20 m` `T=2pisqrt((0.2)/(9.8))` `= 2 xx3.14 sqrt((0.2)/(9.8))=0.897s` |
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| 14432. |
A particle of mass M is situated at the center of a spherical shell of same mass and radius a. The gravitational potential at a point situated at a/2 distance from the centre, will be....... |
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Answer» `-(3GM)/a` = Potential energy at a given point due to shell + potenital energy at given point due to particle `=-(GM)/a+(-(GM)/((a/2)))` `=-(GM)/a - (2GM)/a` `= - (3GM)/a` |
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| 14433. |
The earth is rotating with an angular velocity of 7.3 xx 10^(-5) rad/s. What is the tangential force needed to stop the earth in one year ? Given moment of inertia of the earth about the axis of rotation = 9.3 xx 10^(37) m^(2). Radius of the earth = 6.4 xx 10^(6) m |
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Answer» SOLUTION :Initial angular frequency `=omega_(0) = 7.3 xx 10^(-5)` rad/s Final angular VELOCITY `=omega =0` Time `=t = 1 "year" = 365.25 xx 24 xx 60 xx 60s` Angular acceleration `=alpha`=? `omega = omega_(0) + alpha t` `alpha = (omega - omega_(0))/t = (-7.3 xx 10^(-5))/(36.25 xx 24 xx 60 xx 60) = -2 xx 10^(-12) rad//s^(2)` Torque =`tau =laalpha = F xx R` TANGENTIAL force = F = ? Radius of the earth =`R = 6.4 xx 10^(6) m` M.I. of the earth `= I = 9.3 xx 10^(37) kg m^(2)` `F = (I alpha)/R = (9.3 xx 10^(37) xx 2 xx 10^(-12))/(6.4 xx 10^(6)) = 2.9 xx 10^(19) N` |
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| 14434. |
A particle of mass m is executing oscillations about the origin on X-axis. Its potential energy isU(x) = -K|x|^3, where K is a positive constant. If the amplitude of oscillation is a, then its time period T is: |
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Answer» PROPORTIONAL to `1//SQRT(a)` |
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| 14435. |
A block of wood of mass 0.5 kg is placed on a plane making 30^(@) with the horizontal. If the coefficient of friction between the surfaces of contact of the body and the plane is 0.2. What force is required to keep the body sliding down uniform velocity. |
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Answer» Solution :FORCE required to keep the body SLIDING down with CONSTANT velocity. Given, m = 0.5 KG `g=9.8 ms^(-2)`, `THETA = 30^(@)` `mu = 0.2` `F=mg(sin theta - mu cos theta)` = 1.6 N. |
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| 14436. |
A chain AB of length l is located on a smooth horizontal table so that its fraction of length h hange freely with end B on the table . At a certain moment , the end A of the chain is set free . With what velocity with this end of the chain slip off the table ? |
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Answer» Solution :Let m be the mass per unit length of the chain . Let at any instant X be the length of hanging part of chain and T the tension in chain . Then `x mg -T=x ma`….(1)andT=(l-x)ma…..(2) ADDING these equations , we get `x mg=[XM+(l-x)m]a=l ma` (or) `a=x/lg , As a=(dv)/(dt) rArr (dv)/(dt) =x/lg` Multiplying both sides by `(dx)/(dt) , ` we get `(dx)/(dt) (dv)/(dt) =x/l g (dx)/(dt)"or"vdv= x/lg dx` Integrating both sides, we get `int_(0)^(v) v dv =g/l int_(H)^(l) x dx` `[v^(2)/2]_(0)^(v)= g/l[x^(2)/2]_(h)^(l) =g/l[(l^(2)-h^(2))/2]:. v=sqrt{(g(l^(2)-h^(2)))/l}. |
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| 14437. |
A 5kg collar is attached to a spring of spring constant 500 Nm^(-1). It slides without friction over a horizontal rod. The collar is displaced from its equilibrium position by 10.0 cm and released. Calculate the period of oscillation. |
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Answer» Solution :SPRING constant `K= 500 Nm^(-1)` MASS of collar `m = 5kg` Amplitude `A = 10 CM = 0.1 m` The period of oscillation of collar attached at the end of spring. `T= 2pi sqrt((m)/(k))` `= 2 xx3.14 xx sqrt((5)/(500)) = 6.28 xx (1)/(sqrt(1000))` `=(6.28)/(10) = 0.628 approx 0.63s` |
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| 14438. |
The range of the projectile depends |
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Answer» The ANGLE of projection |
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| 14439. |
Three particles of masses 1 kg, 2 kg and 3 kg are placed at the vertices A, B and C of an equilateral triangle ABC. If A and B lie at (0, 0) and (1, 0) m, the co-ordinates of their centre of mass are |
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Answer» `((sqrt(3))/(2)"m and"(7)/(6)m)` |
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| 14440. |
A particle executing SHM is described by the displacement function x(t)=Acos(omegat+phi), if the initial (t=0) position of the particle is 1 cm, its initial velocity is pi" cm "s^(-1) and its angular frequency is pis^(-1), then the amplitude of its motion is |
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Answer» `picm` At t=0, x=1 cm `therefore1=Acosphi`. . . (i) Velocity, `v=(dx)/(dt)=(d)/(dt)(aCos(omegat+phi))=-Aomegasin(omegat+phi)`. . . (II) SQUARING and adding (i) and (ii), we get `A^(2)cos^(2)phi+A^(2)sin^(2)phi=2` `A^(2)=2""(becausesin^(2)phi+cos^(2)phi=1)` `thereforeA=sqrt(2)cm` |
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| 14441. |
A large block of ice 5mthick has a vertical hole drilled in it and is flouting in a lake, the minimum length of the required to draw a bucketfull of water through the hole is (density of ice 900kg//m^(3)) |
| Answer» ANSWER :A | |
| 14442. |
A particle is placed at the lowest point of a smooth parabola having the equation x^2 = 5y( x,y are in metre). After a slight displacement, the particle is constrained to move along the parabola. Find the angular frequency of oscillations (in rad/sec). |
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Answer» |
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| 14443. |
If surface tension of soap solution is equal to 2.5 xx10^(-2) Nm^(-1) , the work done to blow a soap bubble of diameter 0.6cm against surface tension forces will be |
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Answer» `5.65 xx 10^(-6) J` |
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| 14444. |
There is an obstruction of height h in front of a wheel of radius r weighing W. What is the minimum horizontal force that is to be applied at the centre O of the wheel to overcome the obstruction? Given, h ltr. |
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Answer» Solution :The wheel touches the obstacle at point B FIG. Suppose a force F attempts to turn the wheel about the point B. Weight of wheel W , on the other HAND, resists this ROTATION. Hence to overcome the obstacle MOMENT of the force F about B must be GREATER than the moment of W about B. For F to be minimum, `FxxBD=WxxBC` Here, BD = OC = OA-AC=r-h BC = `sqrt(OB^(2)-OC^(2))=sqrt(r^(2)-(r-h)^(2))=sqrt(2rh-h^(2))` Hence, F (r-h) = W` sqrt(2rh-h^(2))` or, F = `(Wsqrt(2rh-h^(2)))/(r-h)`. 
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| 14445. |
A wire is suspended vertically from one of it.s ends is stretched by attaching a weight of 200N to the lower end. The weight stretches the wire by Imm. Then the elastic energy stored in the wire is |
| Answer» ANSWER :4 | |
| 14446. |
When a force is applied on a body, it can change |
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Answer» velocity |
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| 14447. |
The ratio of dimensions of angular and linear momemtum is |
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Answer» `M^0L^1T^0 ` |
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| 14448. |
A particle is projected at 60^@ to the horizontal with a kinetic energy K. The kinetic energy at the highest point is |
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Answer» K/2 |
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| 14449. |
(A) : A body of mass 10 kg is placed on a rough inclined surface (mu = 0.7) . The surface is inclined to horizontal at angle 30^(@) . Acceleration of the body down the plane will be zero . (R) : Work doneagainst friction always converts to heat . |
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Answer» If both A and R are TRUE and R is the correct explanation of A . |
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| 14450. |
A circular disc of radius 7//picm is placed horizontally on the surface of water. Find the vertical force required to separate it from the water surface. (Surface tension of water = 0.070N//m) |
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Answer» SOLUTION :The downward FORCE on the circumference of the CIRCULAR disc = `Sxx2piR` Where R is the radius of the circular disc The vertical force required to separate it from the water surface = Downward force `F=Sxx2piR` `S=0.070N//m,R=(7)/(pi)CM=(7)/(pi)xx10^(-2)m` `F=0.070xx2xxpixx(7)/(pi)xx10^(-2)=9.8xx10^(-3)N` |
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