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.
| 14301. |
The equation of trajectory of a projectile is y = 10 x - ((5)/(9)) x^(2). If we assume g = 10 ms^(-2) the range of projectile (in meters) is |
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Answer» 36 |
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| 14302. |
Two particles each of mass m separated by a distance d and move in a uniform circle under the action of their mutual force of attraction. The speed of each particle is |
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Answer» `sqrt((Gm)/d)` |
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| 14303. |
The length of a second's pendulum on the surface of the Earth is 0.9m. The length of the same pendulum of surface of planet X such that the acceleration of planet X is n times greater than the Earth is : |
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Answer» `0.9 N` `l=0.9m, T = 2pi sqrt((l_(x))/(g_(x)))` `g_(x)=NG` `l_(x)=0.9 n` |
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| 14304. |
Express giga in terms of femto. |
| Answer» SOLUTION :`1 "giga"=10^(9) a` and `1 "FEMTO"= 10^(-15)a`, where a is a reference standard. So `1 "giga"=10^(9)"/"10^(-15)= 10^(24)` femto. | |
| 14305. |
The figure shown a graph of current in a dicharge circuit of a capacitor through a resistor of resistar 10 Omega. Choose the correct option (s) |
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Answer» The INITIAL potential difference across the capacitor 100 Volts |
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| 14306. |
When a system is taken from state I to state f along thepath iaf, it is found that Q = 50 cal and W = 20 cal. Along the path ibf, Q = 36 cal. W along the path ibf is |
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Answer» Solution :According to first law of thermodynamics, For the path iaf, `Q_(iaf) = DeltaU_(iaf) + W_(iaf)` or `Delta U_(iaf) = Q_(iaf) - W_(iaf) = 50 - 20 = 30 cal` For the path IBF, `Q_(ibf) = Delta U_(ibf) + W_(ibf)` As change in INTERNAL energy is path independent, so `Delta U_(iaf) = Delta U_(ibf) :. Q_(ibf) = Delta U_(iaf) + W_(ibf)` or `W_(ibf) = Q_(ibf) - Delta U_(iaf) = 36-30 = 6 cal` |
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| 14307. |
Passage - II : A long slender rod of mass 2kg and length 4m is placed on a smooth horizontal table. Two particles of massless 2kg and 1kg strike the rod simultaneously and stick to the rod after collosion as shown in Angular velocity of the rod after collision is |
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Answer» RAD /s |
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| 14308. |
Passage - II : A long slender rod of mass 2kg and length 4m is placed on a smooth horizontal table. Two particles of massless 2kg and 1kg strike the rod simultaneously and stick to the rod after collosion as shown in Velocity of the centre of mass of the rod after collision is |
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Answer» 12 m/s |
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| 14309. |
Passage - II : A long slender rod of mass 2kg and length 4m is placed on a smooth horizontal table. Two particles of massless 2kg and 1kg strike the rod simultaneously and stick to the rod after collosion as shown in If the two particles strike the rod in opposite direction, then after collision, as compared to the previous situation the rod will |
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Answer» ROTATE faster but translate at the same rate |
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| 14310. |
By the instrument of 1 kW power, 1 kWh energy is consumed in …... time. |
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Answer» |
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| 14311. |
A hoop of mass 500g and radius 10cm is placed on a nail. Find the moment of inertia of the hoop when it is rotated about the nail. |
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Answer» SOLUTION :When the hoop is rotated about the NAIL, the axis of rotation is the tangent PERPENDICULAR to its plane. Its moment of INERTIA `I=2MR^(2)=2xx0.5xx(0.1)^(2)=0.01kgm^(2)` |
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| 14312. |
A solid sphere and a hollow sphere of the same material and of equal radii are heated to the same temperature |
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Answer» Both will EMIT EQUAL AMOUNT of radiation per UNIT time in the beginning |
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| 14313. |
A vibrating stretched string resonates with a tuning fork of frequency 512 Hz when the length of the string is 0.5 m . The length of the string required to vibrate resonantly with a tuning fork of frequency 256 Hz would be ……… . |
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Answer» 0.25 m |
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| 14314. |
If energy is neither created nor destroyed, what happens to the so much energy spent against friction? |
| Answer» SOLUTION :The energy is dissipated in the form of HEAT. The heat energy so PRODUCED is not AVAILABLE for work. | |
| 14316. |
In a 10 metre deep lake, the bottom is at a constant temperature of 4^(0)C. The air temperature is constant at -4^(0)C. The thermal conductivity of ice is 3 times that of water. Neglecting the expansion of water on freezing, the maximum thickness of ice will be |
| Answer» ANSWER :A | |
| 14317. |
When a gas is supplied 'dQ' heat, it performs a work 'dW' . The increase in its internal energy 'dU' is |
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Answer» DU=dQ+dW |
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| 14318. |
State and prove Kepler's second law (Law of Areas) of planetary motion. |
| Answer» SOLUTION : The LINE that JOINS any PLANET to the SUN sweeps equal areas in equal intervals of time. | |
| 14319. |
The lengths of two steel wires are 100 cm and 200cm and their radii are 6 and 3 mm.nder the same stretching force, what is the ratio of the elongations produced in them. |
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Answer» |
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| 14320. |
The surface over which blocks are placed in smooth. What is the acceleration of each block in the given diagram ? |
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Answer» `9 m//s^(2)` |
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| 14321. |
The Bernoulli equation can be considered to be a statement of the conservation of energy . |
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Answer» |
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| 14322. |
Round offthe following numbers up to 3 digits : (i) 17.65 (ii) 14,958 (iii) 3,49,338 (iv) 11.6 |
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Answer» SOLUTION :(i) `17.6` (II) 15,000 (III) 3,49,000 (iii) 11.6. |
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| 14323. |
If vectors are vecA = 2 hati + 3 hatJ - hatk and hatB = 4 hati + 6 hatJ - 2 hatk , then they are ......... |
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Answer» equal
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| 14324. |
A tube of length closed at one end is lowered into a tank of mercury to a depth H Mercury rises to a height into the tube. It the mercury barometer stands at then |
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Answer» `H=(2H-h))(H-h)` |
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| 14325. |
The equation for the motion of a particle is v = at . The distance travelled by the particle in the first 4 second is : |
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Answer» 8a ` UNDERSET(0)overset(x)intdx = underset(0)overset(4)intat dt ` `x = a [(t^(2))/(2)]_(0)^(2) = (a)/(2) [ (4)^(2) - (0)^(2)]` `x = (a)/(2) xx 16 = 8a` |
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| 14326. |
When a rod is heated but prevented from expanding, the stress developed is independent of |
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Answer» Material of the rod |
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| 14327. |
For a hollow cylinder and a solid cylinder rolling without slipping on a inclined plane, then which of these reaches earlier: |
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Answer» SOLID cylinder |
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| 14328. |
A sample of gas consists of mu_(1) moles of monoatomic molecules, mu_(2) moles of diatomic molecules and mu_(3) moles of linear triatomic molecules. The gas is kept at high temperature. What is the total number of degrees of freedom? |
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Answer» `[3mu_(1)+7(mu_(2)+mu_(3))]N_(A)` |
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| 14329. |
A particle performs simple harmonic motion with speed v and acceleration a. Which of following statement is true ? |
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Answer» Velocity is MAXIMUM, WHEREAS acceleration is zero. `THEREFORE v= A omega` is maximum value of velocity and in acceleration `a= -omega^(2)x, x=0, a=0`. |
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| 14330. |
Consider a frictionless ramp on which a smooth object is made to slide down an initial height 'h'. The distance 'd' necessary to stop the object on a flat track (of coefficient of friction 'mu'), kept at the ramp end is |
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Answer» `H//MU` |
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| 14331. |
A man of mass 40 kg is at rest between the walls as shown in the figure. If 'mu' between the man and the walls is 0.8, find the normal reactions exerted by the walls on the man. |
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Answer» SOLUTION :SINCE MAN is at rest, `N_(1)-N_(2)=0` `therefore N_(1)=N_(2)=N` (SAY) `therefore N_(1)=N_(2)=N` (say) `therefore 2mu N=mg` `= 2xx0.8xx N=400` `therefore N = 250 N`
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| 14332. |
A body undergoes no change in volume. Poisson's ratio is |
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Answer» 0.5 or `(Delta V)/(V) = (Delta l)/(l) + (2 Delta r)/(r )` As `Delta V= 0` (Given), `0 = (Delta l)/(l)(2 Delta r )/(r )` or `(Delta l)/(l) = - (2DELTA r)/(r )` Poisson.s ratio, `sigma = - (Delta r//r)/(Delta l//l) = (Delta r//r)/(2 Delta r//r) = (1)/(2) = 0.5` |
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| 14333. |
A bullet fired from a rifle attains a maximum height of 5 m and crosses a range of 200 m. Find the angle of projection. |
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Answer» |
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| 14334. |
The resistance R of a metallic conductor is equal to the ratio of potential difference V across the resistor to the current I through the conductor. Calculate R given tat V=(11.1+-0.1) volt and I=(5.5+-0.5)A |
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Answer» `2.2=(0.1//l)100+(0.1//7.9)100`, SOLVING l=10.6 CM. |
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| 14335. |
(A) : The change in kinetic energy of a particle is equal to the work done on it by the net force.(R ) : Change in kinetic energy of particle is equal to the work done only in case of a system of one particle. |
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Answer» Both (A) and (R ) are TRUE and (R ) is the correct EXPLANATION of (A) |
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| 14336. |
(A): Body projected vertically up or down from the top of a tower with same velocity will reach the ground with same velocity. (R) : Both the bodies projected vertically up and down from the top of the tower will have same. displacement and acceleration. |
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Answer» Both (A) and (R) are TRUE and (R) is the CORRECT explanation of (A) |
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| 14337. |
End correction in a closed organ pipe of diameter 'd' is |
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Answer» `0.6d` |
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| 14338. |
Two pendulums of lengths 100 cm and 225 cm start oscillating in phase simultaneously. After how many oscillations will they again be in phase together ? |
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Answer» |
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| 14339. |
The diagram shows the profile of a wave, which of the followingpairs of points are in phase? |
| Answer» SOLUTION :B, E | |
| 14340. |
A solid is completely immersed in a liquid. The force exerted by the liquid on the solid will a) Increase if it is pushed deeper insider the liquid b) Change if its orientation is changed c) Decrease if it is taken partically out of the liquid d) Be in the vertically upward direction |
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Answer» a, and C are correct |
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| 14341. |
These both situation is truely shown in graph (D) A satellite of mass m is orbiting the earth (of radius R) at a height h from its surface. The total energy of the satellite in terms of g_0 , the value of acceleration due to gravity at the earth's surface, is |
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Answer» `(2mg_0R^2)/(R+h)` `U = -(GM_em)/(2r)` `:. U = - (GM_e_m)/(2(R+h)) ""...(1) "" [ :. r = R + h]` Now `g_0 = (GM_e)/R^2` `:. GM_e = g_0R^2` In equ. (1) `GM_e = g_0 R^2` `U = -(mg_0R^2)/(2(R+h))` |
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| 14342. |
How much heat energy in joules must be supplied to 14 gms of nitrogen at room temperature to raise its temperature by 40^(@)C at constant pressure. |
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Answer» 50R |
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| 14343. |
A body is sliding down a rough inclined plane. The coefficient of friction between the body and the plane is 0.5. The ratio of the net force required for the body to slide down and the normal reaction on the body is 1:2. Then the angle of the inclined plane is |
| Answer» Answer :C | |
| 14344. |
The coefficients of static and dynamic friction are 0.7 and 0.4. The minimum force required to create motion is applied on a body and if it is further continued, the acceleration attained by the body in ms^(-2) is (g = 10m//s^(-2)) |
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Answer» 7 |
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| 14345. |
An object of mass 1 kg is falling from the heighth - 10 m. Calculate (a) The total energy of an object at h = 10 (b) Potential energy of the object when it is at h = 4 m (c) Kinetic energy of the object when it is at h = 4 m (d) What will be the speed of the object when it hits the ground? Assume g = 10 ms^(-2)) |
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Answer» Solution :(a) The gravitational force is a conservative force. So the total energy remains constant throughout the motion. At h = 10 m, the total energy E is entirely potential energy `R=U=mgh=1xx10xx10=100J` (b) The potential energy of the object at h = 4 m is `U=-mgh=1xx10xx4=40J` (c) Since the total energy is constant throughout the motion, the kinetic energy at h= 4m must be `KE=E- U=100-40=60j` ALTERNATIVELY, the kinetic energy COULD also be found from velocity of the object at 4 m. At the height 4 m, the object has fallen through a height of 6 m. The velocity after falling 6 m is calculated from the EQUATION of motion, `v=sqrt(2gh)=sqrt(2xx10xx6)=sqrt(120)ms^(-1)` The kinetic energy is `KE=1/2mv^(2)=1/2xx1xx120=60j` (d) When the object is just about to hit the ground, the total energy is completely kinetic and the potential energy, U= 0. `E=KE=1/2mv^(2)=100J` `v=sqrt(2/mKE)=sqrt(2/1xx100)=sqrt(200)=14.12ms^(-1)` |
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| 14346. |
Assuming that the moon completes one revolution in a circular orbit around the earth in 27.3 days, calculate the acceleration of the moon towards the earth. The radius of the circular orbit can be taken as 3.85 xx 10^(5)km. |
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| 14347. |
Compare the components for the following vector equations(a)That(j) - mg hat(j) = ma hat(j),(b) vec(T) + vec(F) = vec(A) +vec(B)(c)vec(T) - vec(F) = vec(A) - vec(B),(d)T hat(j) +mg hat(j) = ma hat(j) |
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Answer» Solution :(a) T - mg = ma (B) `T_(x) +F_(x) = A_(x) +B_(x)` (C)`T_(x) - F_(x) = A_(x) - B_(x)` (d)T + mg = ma |
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| 14348. |
A dynamometer D is attached to two bodies of masses M=6 kg and m=4 kg. Force F=20 N and f=10N are applied to the masses as shown. The dynometer reads |
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Answer» 10N `a=(F-f)/(M+m)=(20-10)/(6+4)=1 ms^(-2)"" ` (TOWARDS left) Let `F_(0)` be the reading of DYNAMOMETER, then equation of motion of MASS m would be, `implies F_(0)=f+ma=10+(4)(1)=14N` |
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| 14349. |
A circular disc is rotating about its own axis at a uniform angular velocity . omega.The disc is subjected to uniform angular retardation by which its angular oo velocity is decreased to omega/2during 120 rotations. The number of rotations further made by it before coming to rest is: |
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Answer» 120 |
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| 14350. |
A spherical soap bubble (surface tension S) encloses n moles of a monoatomic ideal gas the gas is heated slowly so that the effective surface area of the bubble increases by (3nR)/(2S)per unit incrementin temperature of the gas. The specific heat for this process is (Assume surrounding is vacuum) |
| Answer» Answer :D | |