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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.
| 15101. |
Force in Column -I and its use is in Column -II are given .Match them appropriately. |
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| 15102. |
A uniform plate of mass m is suspended in each of the ways shown. For each case determine immediately after the connection at Bhas been released , (a) the angular acceleration of the plate. (b) the acceleration of its mass center. |
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Answer» `I_(Z Z') = I_(XX') + I_(YY')` `I_(Z Z') = (m(c//2)^(2))/(12) +(mc^(2))/(12) rArr I_(Z Z') = (5mc^(2))/(48)` from parallel axis theorem `I_(A) = I_(Z Z') + mr^(2)` `I_(A) = (5 mc^(2))/(48) + m(( c)/(4) SQRT(5))^(2)` `I_(A) = (5mc^(2))/(48) + (m xx 5c^(2))/(16)` `I_(A) = (5mc^(2))/(12)` (i) As 'B' is RELEASED (a) `tau = I ALPHA` `mg xx (c)/(2) = ((5mc^(2))/(12)) alphaalpha = (1.2 g)/(c)` (b) `a_(CM) =a_(t) cos theta (- hati) + a_(t) sin theta (-hat j) and a_(t) = r alpha` `a_(CM) = alpha (r costheta) (-hat i) + alpha r sin theta (-hat j)` `a_(CM) = alpha(c)/(4) (-hat i) + prop (c)/(2) (-hatj) rArr a_(CM) = -(alphac)/(4) [hat i + 2 hatj]` `a_(CM) = -0.3[hat i + 2 hat j] g` (b) `2 F = mg F = (mg)/(2)` (spring force) As the spring at `B` is cut for transverse motion `mg - F = ma` `(mg)/(2) = ma a = (g)/(2) darr` for rotational motion with resect to `A` `(mg -ma) ( c)/(2) = I alpha` (`ma` in pssudo force) `(mg)/(2) xx ( c)/(2) = ((5 mc^(2))/(12) alpha` `alpha = (2.4 g)/( c) (c.w.)`.   .
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| 15103. |
T_(1), T_(2)and T_(3)aretensionsin thethreestrings.ThenT_(1): T_(2)is(##CHY_DMB_PHY_XI_P2_QP_19_E01_005_Q01.png" width="80%"> |
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Answer» `SQRT(3): 2` |
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| 15104. |
Dimension of force and stress are equal. |
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Answer» `[STRESS]=[M^1L^-1T^-2]` |
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| 15105. |
The bob of a pendulum of length 2 m lies at P. When it reaches Q it loses 10% of its total energy due to air resistance. For this event, which of the following is a correct statement. |
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Answer» The velocity at Q is `1m//s` K.E. at Q = 90% of P.E. at P `(1)/(2)mv^(2)=(90)/(100)mgh` `v^(2)=0.9ghxx2` `v=sqrt(1.8gh)` Here `g=10m//s^(2) h=2` `THEREFORE` Velocity at `Q,v=sqrt(1.8xx10xx2)=sqrt(36)=6m//s` The other statements are incorrect. |
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| 15106. |
A 70 kg man stands in contact against the inner wall of a hollow cylindrical drum of radius 3 m rotating about its vertical axis 200 rev/min. The coefficient of friction between the wall and his clothing is 0.15. What is the minimum rotational speed of the cylinder to enable the man to remain stuck to the wall (without falling when the floor is suddenly removed ? |
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Answer» Solution :The frictional force f (vertically upwards) opposes the weight mg. The man remains stuck to the WALL after the floor is removed if `mg LE f_(L)` i.e., `mg lt mu m R omega^(2)`. The MINIMUM angular speed of rotation of the cylinder is `omega_(min)=sqrt((G)/(mu R))=sqrt((9.8)/(0.15xx3))=4.67 s^(-1)`. |
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| 15107. |
The dimension of the power is |
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Answer» `ML^(2)T^(-2)` |
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| 15108. |
The time of oscillation of a small drop of liquid under surface tension depends upon density rho, radius r and surface tension s, as T alpha rho^(a) s^(b) r^( c), then the descending order of a, b and c is |
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Answer» `a GT B gt C` |
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| 15109. |
Two identical spheres move in opposite directions with speeds v_(1) and v_(2) and pass behind an opaquescreen, where they may either cross without touching (Event 1) or make an elastic head-on collision (event 2) |
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| 15110. |
Find wave speed in a string of length 7m, mass 0.035 kg and having a tension of 60.5 N. |
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Answer» (A) 77 m/s` `therefore v = sqrt (( 60.5 xx7 )/(0.035)) =110 (m)/(s)` |
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| 15111. |
A force of 2 hat(i) + 3hat(j) + 2hat(k)N acts a body for 4s and produces a displacement of 3hat(i) + 4hat(j) + 5hat(k)m calculate the power? |
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Answer» 5W |
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| 15112. |
An elastic spring of unstretched length L and force constant k is stretched by a small length x. It is further stretched by another small length y. The work done in the second stretching is |
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Answer» `1/2 KY^2` `:.`WORK done = `U_2 - U_1 = 1/2 k(x + y)^2 - 1/2 kx^2` `= 1/2 ky (2x + y)`. |
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| 15113. |
A body is rolling down an inclined plane. If K.E. of rotation is 40% of K.E. of translation, then the body is a |
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Answer» SOLID sphere |
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| 15114. |
Find the centre of mass of uniform thin sheet as shown in figure. |
| Answer» SOLUTION :`(9, 15)` | |
| 15115. |
Find the centre of mass of uniform thin sheet as shown in figure. |
| Answer» SOLUTION :`(8.33, 15)` | |
| 15116. |
A mass of 2 kg is rotating on a circular path of radius 0.8 m with angular velocity of 44 rad s^(-1). If the radius of the path becauses 1.0 m, what will be the value of angular velocity? |
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| 15117. |
In a streamline flow |
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Answer» the speed of a partice always REMAIN same |
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| 15118. |
("he")/(4pi m) has the same dimensions as (h= Planck's constant, e= charge, m= mass) |
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Answer» MAGNETIC moment |
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| 15119. |
What are thermal radiations and black body radiations ? |
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Answer» SOLUTION :Thermal RADIATIONS: The energy emitted in the from of radiation by a BODY die to its temperature is called thermal radiation. Black body radiation:The radiation emitted by a body place in an isothermal ENCLOSURE in equilibrium is called black body radiation. |
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| 15120. |
Look at the graphs (a) to (d) (Fig. 3.20) carefully and state, with reasons, which of these cannot possibly represent onedimensional motion of a particle. |
Answer» SOLUTION : Here, none of the graphs dosen.t represent the direction of motion in one DIMENSION. (a) Two positions are there at same time. Hence, it is not for motion in one dimension. (b) There are positive and negative values of velocity for same time. It is not possible for motion in one dimension. (c) Negative value of speed is not possible because it is ALWAYS positive. Hence it is not for motion in one dimension. (d) Path length cannot decrease with RESPECT to time during motion. Hence, it is also not for motion in one dimension. Note that the DIRECTIONS of arrows in graph are meaningless. |
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| 15121. |
The kinetic energy of a body is .K.. If one - fourth of its mas is removed and velocity is doubledm its new kinetic energy is, |
| Answer» ANSWER :B | |
| 15122. |
Explain why the water doesn't fall even at the top of the circle when the bucket full of water is upside down rotating in a vertical circle? |
| Answer» Solution :WEIGHT of the water and bucket is used up in PROVIDING the NECESSARY CENTRIPETAL force at the top of the CIRCLE | |
| 15123. |
If n _(1), n _(2), and n _(3)are the fundamental frequencies of three segments into which a string is divided, then the original fundamental frequency n of the string is given by … |
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Answer» `(1)/(n) = (1)/( n _(1)) + (1)/( n _(2)) + (1)/(n _(3))` (1) For FIRST PART of length `L_(1) , n _(1) = (1)/(2L_(1)) sqrt((T)/(mu))` (2) For second part of length `l _(2), n _(2) = (1)/(2L _(2)) sqrt ((T)/( mu))` (3) For the part of length `L_(3), n_(3) = (1)/(2L _(3)) sqrt ((T)/(mu))` But `L = L _(1) + L _(2) + L _(3)` `therefore (1)/(2N ) sqrt ((T)/( mu)) = (1)/( 2n _(1)) sqrt ((T)/( mu ))+ (1)/( 2n _(2)) sqrt ((T)/( mu )) + (1)/( 2n _(3)) sqrt ((T)/( mu ))` `therefore (1)/(n) = (1)/( n _(1) )+ (1)/(n _(2)) + (1)/( n _(3))` |
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| 15124. |
Among the following, which of the forces are long range? (i) Force between two nucleons, (ii) Force betweentwo electric charges, (iii) Force between the earth and the moon. |
| Answer» SOLUTION :Gravitational FORCE between earth and MOON is the long RANGE force. | |
| 15125. |
In a spring gun having spring constant 100N/m, a small ball of mass 0.1kg is put in its barrel by compressing the spring through 0.05m as shown in Figure. The ball leaves the gun horizontally at a height of 2m above the ground. Find (a) The velocity of the ball when the spring is released. (b) Where should a box be placed on the ground so that the ball falls in it. (g= 10 m//s^(2)) |
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Answer» Solution :(a) When the spring is released its elastic potential energy `(1//2) kx^(2)` is converted into kinetic energy `(1//2) mv^(2)` of the ball, so, by conservation of MECHANICAL energy `(1)/(2) mv^(2)= (1)/(2) kx^(2) rArr V= x SQRT((k)/(m))` So `v= 0.05 sqrt((100)/(0.1)) m//s = sqrt((5)/(2)) m//s` (b) As initial vertical component of velocity of ball is zero, time taken by the ball to reach the ground.  `t= sqrt((2h)/(g))= sqrt((2 xx 2)/(10))= sqrt((2)/(5)) s`, So the horizontal distance TRAVELLED by the ball in this time `d= vt = sqrt((5)/(2)) xx sqrt((2)/(5))= 1m` |
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| 15126. |
A barometer kept in a elevator accelerating upwards with acceleration a . Fing most likely pressure inside the elevator. |
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Answer» SOLUTION :The RESULTANT acceleration of elevator in upwards DIRECTION =a+g `therefore` The pressure in elevator `=HRHO(g+a)` `(76xx13.6xx(g+a))/(13.6xxg)`cmHg This pressure is greater than the atmospheric pressure 76 cm Hg . |
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| 15127. |
Escape velocity of a body of 1 kg. On a planet is 100 ms^(-1). Gravitational potential energy of the body at the planet is ......... |
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Answer» `-5000` J ` :.P.E ., U = -(GMm)/2 =-5000 J ` |
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| 15128. |
A cyclotron accelerates a proton to a final speed of 3 xx 10^(7) ms^(-1) which is initially at rest. Find how much work is done on the proton by the electrical force of the cyclotron. Mass of the proton is 1.67 xx 19^(-27) kg. |
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Answer» Solution :Mass of proton `m = 1.67 xx 10^(-27)` kg , INITIAL velocity of proton u = 0 FINAL velocity of proton `v = 3 xx 10^(7) ms^(-1)` WORK done by the force on the proton = Kinetic energy gained `therefore` Work done `W = (1)/(2) mv^(2) = (1)/(2) xx 1.67 xx 10^(-27) xx (3 xx 10^(7))^(2) therefore W = 7.515 xx 10^(-13)` J |
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| 15129. |
A glass full of hot milk is poured on the table. It begins to cool gradually. correct of the following. a) The rate of cooling is constant till milk at tains the temperature of the surrounding. b) The temperature of milk falls off exponentially with time c) While cooling there is a flow of heat from milk to the surrounding as well as from surrounding to the milk but the net flow of heat is from milk to the surrounding and that is why it cools d) All three phenomenon, conduction, convection and radiation are responsible for the loss of heat from milk to the surroundings. |
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Answer» a and b |
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| 15130. |
You accelerate your car forward. What is the direction of the frictional force on a package resting on the floor of the car? |
| Answer» Solution :The package in the ACCELERATED car (a non-inertial FRAME) EXPERIENCES a Pseudo force in a direction opposite to that of the motion of the car. The FRICTIONAL force on the package which acts opposite to this pseudo force is thus in the same direction (forward) as that of the car. | |
| 15131. |
What is effect of temperature on elasticity ? |
| Answer» SOLUTION :As the temperature of SUBSTANCE INCREASES, its elasticity DECREASES. | |
| 15132. |
Discuss the observation of Galileo for the objects falling freely. |
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Answer» SOLUTION :GALILEO CONCLUDED that the rate of change of velocity with time is a constant of MOTION for all objects in free fall and the change in velocity with DISTANCE is not constant. It decreases with the increasing distance of fall. |
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| 15133. |
The mass of an athlete is 40kg. He can cover a distance of 150m in maximum of 15sec and minimum of 10sec then the range of his kinetic energy is |
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Answer» `2000J-4500J` |
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| 15134. |
A particle of mass 0.1 kgmoving with an initial speed v collide with another particle of same mass kept at rest. If after collision total energy becomes 0.2 J. Then |
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Answer» MINIMUM VALUE of V is 2m/s |
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| 15135. |
Two metallic spheres S_(1)andS_(2) made of same material have identical surface finish. The mass of S, is 3 times that of S_(1) is 3 times that of S_(2). Both are heated to same temperature and are placed in same surroundings. Then ratio of their initial rates of fall of temperature will be |
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Answer» `1/sqrt3` |
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| 15136. |
The mechanical energy (E) of a body is the sum of kinetic energy (K) and potential energy (V) of the body i.e., E = K + V. Whereas K is always positive, V can be positive or negative. For a system to axis, K = (E - V) ge 0 Negative value of E indicates a bound state. For example, all planets revolving around the sum have nagative mechanical energy. Read the above passage and answer the following questions : (i) When mechanical energy E = 0, does it mean K = 0 and V = 0 ? (ii) What is the impulication of yjis study in day to day life ? |
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Answer» Solution :(a) No, not always. This is because, `K` is always positive, so when `V` is equally NEGATIVE, `E = (K + V)` becomes ZERO. (ii) All planet REVOLVING around the sun have negative mechanical energy. It indicates that the planets are bound to revolve around the sun. They are not free to leave their ORBIT. In day to day life, we find that a person is not free to leave the country, if he OWES some amount to income tax (sales tax, etc.) or he has some criminal court cases pending against him. |
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| 15137. |
A rectangular block has a square base measuring axxa , and its height is h, It moves with a speed v on a smooth horizontal surface |
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Answer» It will TOPPLE if `V gt sqrt(2gh)` |
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| 15138. |
If the atmospheric pressure is 76 cn og Hg at what dept of waterin take the pressure will becomes 2 atmospheres nearly |
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Answer» 862 cm |
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| 15139. |
The mass of a particle is 1kg and it is moving along x-axis. The period of its small oscillation is pi/2. Find its potential energy: |
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Answer» `- SIN 2x` |
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| 15140. |
A constant torque of 20 Nm is exerted bri a pivoted wheel for 10 seconds during which" time' the angular velocity of the wheel changes from 0 to 100 rpm. When the external torque is removed'it is stopped by friction in 100 s. Find (i) the MX of the wheel and (ii) the fiictional torque ? |
| Answer» SOLUTION :`alpha = omega//t =2PI XX (5//3)//10 = 1.047 "rads"^(-2), I =tau//alpha = 20//1.047 = 19.10 kg m^(2)` , `a.=0 -10.47//10 =-0.1047 rad s^(-1), tau. = Ia. = -0.1047 xx 19.10 = -2 NM` | |
| 15141. |
A bob attached to one end of a string, other end of which is fixed at peg A. The bob is taken to a position where string makes an angle of to a position where string makes an angle of 30^@ with the horizontal. On the circular path of the bob in vertical plane there is a ped 'B' at a symmetrical position with respect to the position of release as shown in the figure. If V_(c) and V_(a) be the minimum speeds is clockwise and anticlock wise directions respectively, given to the bob in order to hit the ped 'B' then ratio V_(c) : V_(a) is equal to . |
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Answer» `1 : 1` APPLYING CONSERVATION energy `v_(a) = sqrt(2gR)` For clock wise motion `T +mg cos 60 = (mv_(c)^(2))/(R)` `v_(C)` to minimum `T=0 , v_(C) = sqrt((gR)/(2))`. |
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| 15142. |
In the following situation which one of the following is a correct statement? A body projected vertically from the earth reaches a height equal to earth's radius before turning to the earth. The power exerted by the gravitational force is greatest: |
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Answer» at the HIGHEST POSITION of the BODY |
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| 15143. |
A particle performs SHM of time period T. Find the time taken by the particle to go directly (i) from its mean position to half the amplitude and (ii) from half the amplitude position to its positive extreme position. |
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Answer» Solution :When a particle is INITIALLY in the mean position, `X= ASIN omegat` i) When the particle travels from mean position to half its amplitude, DISPLACEMENT, `x = A//2`. `(A)/(2)= Asinomegat` `sinomegat= (1)/(2) implies omegat= (PI)/(6) implies (2pi)/(T)t=(pi)/(6) implies t= (T)/(12)` ii) The time taken to go from half the amplitude position to its positive extreme position = [Time taken to go from mean to positive extreme position]-[Time taken to go from mean to half amplitude position] `= (T)/(4)- (T)/(12)= (T)/(6)` |
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| 15144. |
Two blocks of masses m and M are placed on a horizontal frictionless table connected by light spring as shown in the figure. Mass M is pulled to the right with a force F. If the acceleration of mass m is a, then the acceleration of mass M will be |
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Answer» `((F-ma))/(M)` `F^(1)=ma` For mass .M. `F-F^(1)=Ma^(1)` |
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| 15145. |
Two bodies with K.E in ratio 4:1 are moving with equal linear momentum.The ratio of their masses is |
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Answer» `1:2` |
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| 15146. |
The distance travelled by a moving body is directly proportional to time. Is any external force acting on it? |
| Answer» Solution : As `s PROP t ` , so acceleration a = 0, therefore, no EXTERNAL force is acting on the BODY. | |
| 15147. |
If two vectors overset-A and overset-B are such that |oversettoA+oversettoB|=|oversettoA-oversettoB|, then the angle between the vectors is |
| Answer» ANSWER :C | |
| 15148. |
The ratio of the acceleration for a solid sphere (mass .m. and radius .R.) rolling down and incline of angle .theta. without slipping and slipping down the incline without rolling is |
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Answer» `2:3` |
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| 15149. |
What are the examples of projectile motion? |
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Answer» Solution :An object dropped from WINDOW of a MOVING TRAIN. (ii) A bullet fired from a rifle (iii) A ball thrown in any direction. (iv) A javelin or shot put thrown by an athlete. (v) A jet of water ISSUING from a hole near the bottom of a water TANK. |
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| 15150. |
State the principle of homogeneity. Test the dimensional homogeneity of equations - (i) s = ut + 1/2 at^(2) (ii) S_(n) = u + a/2 (2n - 1) |
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Answer» Solution :(i) DIMENSION of L.H.S [s] = `[M^(0)L^(1)T^(0)]` Dimension of R.H.S. = `[ut] + [at^(2)]` = `[LT^(-1).T] + [M^(0)L^(1)T^(-2).T] = [M^(0)L^(1)T^(0)]` as Dimension of L.H.S = Dimension of R.H.S `:.` The equation is DIMENSIONALLY homogeneous. (ii) `S_(n) =` Distance TRAVELLED in `n^(th)` sec that is `(S_(n) - S_(n-1))` `:.S_(n) = U xx 1 + a/2 [2N - 1]` `[LT^(-1)] = [LT^(-1)] + [LT^(-2)][T]` `[LT^(-1)] = [LT^(-2)]` L.H.S. = R.H.S. Hence this is dimensionally correct. |
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