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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.
| 14151. |
If x, v and a denote the displacement, the velocity and the acceleration of a particle executing simple harmonic motion of time period T, show that the expressions (aT)/xanda^(2)T^(2)+4pi^(2)v^(2) do not change with time. |
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Answer» Solution :If A is the AMPLITUDE, `omega` is the angular frequency and x is the displacement of a particle executing SHM then, velocity `v=pmomegasqrt(A^(2)-x^(2))` acceleration `a=-omega^(2)x` time PERIOD `T=(2PI)/omega` `(aT)/x=(-omega^(2)x*(2pi)/omega)/x=-2piomega` = constant `a^(2)T^(2)+4pi^(2)v^(2)=(-omega^(2)x)^(2)((2pi)/omega)^(2)+4pi^(2)omega^(2)(A^(2)-x^(2))` `=omega^(4)x^(2)(4pi^(2))/omega^(2)+4pi^(2)omega^(2)A^(2)-4pi^(2)omega^(2)x^(2)` `=4pi^(2)omega^(2)A^(2)` = constant |
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| 14152. |
A fish rising vertically to the surface of water in a lake uniformly at the rate of 3ms^-1 observes a bird diving vertically towards the water at a rate of 9ms^-1 vertically above it. If the refractive index of water is 4//3, the actual velocity of the dive of the bird is |
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Answer» 6 |
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| 14154. |
Centre of mass of the earth-moon system lies |
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Answer» CLOSER to the EARTH |
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| 14155. |
When a mass is rotating in a plane about a fixed point, it angular momentum is directed along |
| Answer» Solution :The FORCE acting on comet is radial. For radial force TORQUE is ZERO. i.e. angular MOMENTUM is CONSTANT. | |
| 14156. |
There have been suggestions that the value of the gravitational constant G becomes smaller when considered over very large time period (in billions of years) in the future. If that happens, for our earth, |
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Answer» nothing will change If G decreases with time, then gravitational force `F_G`will become WEAKER with time. |
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| 14157. |
Select the correct pair from the following given pair. |
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Answer» Oscillations of a simple PENDULUM and electromagnetic oscillations in tank circuit. |
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| 14158. |
A physicalquantity u is given by the relation u=(B^(2))/(2mu_(0)) Here B= magnetic field strength mu_(0)= magnetic permeability of vacuum The name of physical quantity u is |
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Answer» ENERGY Unit of `u=((N//Am)^(2))/(N//A^(2))=(N^(2)A^(2))/(NA^(2))m^(2))=N/(m^(2))=(NM)/(m^(3))=J/(m^(3))` `=` energy per unit volume =energy density |
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| 14159. |
Which of the following is a non - conservative force ? |
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Answer» electrostatic |
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| 14160. |
Which of the following four statements is false. |
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Answer» A body can have zero VECLOCITY and still be accelerated. |
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| 14161. |
What are the basic requirements of a cooking utensil in respect of specific heat, thermal conductivity and coefficient of expansion? |
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| 14162. |
Two identical rods are joined to form an .X., The smaller angle between the rods is .e.. The moment of inertia of the system about and axis passing through the point of intersection of the rods and perpendicular to their plane is |
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Answer» `ALPHA theta` |
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| 14163. |
For the wave described in exercise 27 plot the displacement (y) versus (t) graphs for x = 0, 2 and 4 m. What are the shaps of these graphs ? In which aspects does the oscillatory motion in travelling wave differ from one point to another amplitude, frequency or phase? |
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| 14164. |
What happens when a capillary tube of insufficient length is dipped in a liquid? Will the liquid overflow? |
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Answer» Solution :We have `H=(2S0(Rrhog)` where R is the radius of curvature of the liquid meniscus. `HR=(2S)/(rrhog)` = a constant. So hR = h.R. Hence in a capillary tube of insufficient LENGTH, the liquid RISES to the top and spresds out to a new radius of curvature R. given by R. = hR/h.. The liquid will not overflow. |
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| 14165. |
If a spring constant k is divided into n equal parts, then the spring constant of each parth is : |
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Answer» `((N)/(n+1))^(K)` |
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| 14166. |
When a force of 80 N is applied on a uniform steel wire, its free end elongates by 1.2 mm. Find the energy stored in the wire. |
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| 14167. |
A particle has displacementx given by x = 3 cos(5pi t + pi), where x is in metres and in seconds. Where is the particle at time t=0? At time t= 1/2s? |
| Answer» SOLUTION :`X(0) = - 3 m , x(1/2) = 0 ` | |
| 14168. |
A fly wheel of moment of inertia 3 xx 10^(2) kg mạis rotating with uniform angular speed of 4.6 rad s^(-1). If a torque of 6.9 xx 10^2 N m retards he wheel, then the time in which the wheel comes to rest is |
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Answer» 1.5 s TORQUE `tau = 6.9 xx 10^(2)` N m Initial angular SPEED `omega_(0) = 4.6 rad s^(-1)` Final angular speed , `omega = 0 rad s^(-1)` As `omega = omega_(0) + alpha t` `therefore alpha = (omega - omega_(0))/(t) = (0 - 4.6)/(t) = - (4.6)/(t) rad s^(-2)` Negative sign is for DECELERATION . Torque , `tau = I alpha` `6.9 xx 10^(2) = 3 xx 10^(2) xx (4.6)/(t) or t = (3 xx 10^(2) xx 4.6)/(6.9 xx 10^(2)) = 2 s` |
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| 14169. |
A glass tube of uniform internal radius (r ) has a valve separating the two identical ends. Initially, the valve is in a tightly closed position. End 1 has a hemispherical soap bubble of radius r. End 2 has sub-hemispherical sop bubble as shown in figure. Just after opening the valve, |
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Answer» AIR from end 1 flows TOWARDS end 2. No change in the volume of the soap bubbles |
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| 14170. |
A capillary tube of length 5 cm and diameter 1 mm is connected to a tank horizontally. The rate of flow of water is 10 ce per minute. Calculate the rate of flow of water through another capillary tube of diameter 2 mm and length 50 cm is connected in series with the first capillary? |
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Answer» When in SERIES ` P= P_1 +P_2 = ( 8 eta l_1Q_1//pi r_1^(4))+(8 eta l_2 Q_2) //pi r_2^(4)` SIMPLIFYING `Q_2 = 1.02 xx 10 ^(-7)m^(3)//s` |
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| 14171. |
Match the following {:(,"List -I",,"List - II"),((a),"Elastic collision",(e ),e=0),((b),"Inelastic collision",(f),0lt e lt 1),((c ),"Explosion",(g),e=1),((d),"Plastic collision",(h),"Final K.E." gt "Initial K.E."):} |
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Answer» a-e, b-f, c-g, d-h |
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| 14172. |
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.(g=10ms^(-2)) |
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Answer» 100 N |
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| 14173. |
From where do we get highly accurate time signals in U.S? |
| Answer» Solution :Highly accurate TIME signals are obtained from the U.S. Naval Observatory Master Clock. When we have a radio capable of receiving 5, 10 or 15 MHz signals we can pick up co-ordinated UNIVERSAL time signals COMING from National institute of STANDARDS and Technology (NIST) station in Boulder, Colorado, USA. | |
| 14174. |
Consider a two-particle system with the particles having masses m_(1) and m_(2). If the first particle is pushed towards the centre of mass through a distance d. by what distance should the second particle be moved so as to keep the centre of mass at the same position? |
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Answer» `d.=(m_(1))/(m_(2))d` |
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| 14175. |
A body of mass 20 kg is suspended using two ropes. One of the ropes is 30^(@) to the vertical. What should be the inclination of the other rope with the vertical so that tension in this will be minimum ? Find the values of the tensions in the two strings also. |
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| 14176. |
Two flasks R and S of volume V_(1)andV_(2) contain same gas at pressure P_(1)andP_(2) respectively at the same temperature. Pressure of the gas when the flasks R and S are connected by a tube of negligible volume is |
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Answer» <P>`(P_(1)V_(1)+P_(2)V_(2))/(V_(1)+V_(2))` `n_(1)=(P_(1)V_(1))/(RT)andn_(2)=(P_(2)V_(2))/(RT)because N=n_(1)+n_(2)` `therefore P=((n_(1)+n_(2))RT)/(V_(1)+V_(2))=(P_(1)V_(1)+P_(2)V_(2))/(V_(1)+V_(2))` |
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| 14177. |
between two rigid support. The point where the string has to be pluked andtouched are |
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Answer» pluked at `l/4` and TOUCH at `l/2` |
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| 14178. |
The first & second stage of two stage rocket separately weigh 100 kg and 10 kg and contain 800 kg and 90 kg fuel respectively. If the exhaust velocity of gases is 2 km/sec then find velocity of rocket (nearly) (log_(10)5=0.6990) (neglect gravity). |
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Answer» Solution :`V=V_(0)+2.3 U LOG (m_(0)//m)` Velocity as the first stage is detached `= V_(0)` `V_(0)=2.3 ulog(m_(0)//m)` `= 2.3xx2xx10^(3)xxlog (1000//200)` `= 2.3xx2xx0.699xx10^(3)=3.2xx10^(3)` Velocity ACQUIRED with second stage = V `V=V_(0)+2.3u log(m_(0)//m)` `= 3.2xx10^(3)+2.3xx2xx10^(3)xxlog(100//10)` `= 3.2xx10^(3)+4.6xx10^(3)=7.8xx10^(3)m//s` |
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| 14179. |
(a) In which direction should the motorboat given in Illustration 5.46 head in order to reach a point on the opposite bank directly east from the starting point ? The boat's speed relative to the water remains 4 ms^-1. (b) What is the velocity of the boat relative to the earth ? ( c) How much time is required to cross the river ? |
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Answer» SOLUTION :As the boat has to each exact opposite end to the point of start, the boat has to start (velocity `4 m s^-1`) at an angle `theta` aiming somewhat upstream. Taking into count the PUSH given by the current, Velocity of boat w.r.t. river, `vec v_(b,r) = vec(OA) = 4 m s^-1` Velocity of river w.r.t. Earth, `vec v_r = vec(AB) = 2 m s^-1` velocity of boat w.r.t. Earth, `vec v_b m s^-1 = vec(OB) = ?` (a) To find the direction of the boat in which the boat has to GO,we need to find angle `theta`. From `Delta OBA, sin theta = (AB)/(OA) = (2)/(4) = (1)/(2) rArr theta = 30^@` Hence, the motorboat has to head at `30^@` north of east. (b) To find the velocity of boat w.r.t. Earth, we can use pythagorous theorem again. FRO `Delta OBA`, we have `v_(b,r)^2 = v_b^2 + v_r^2 rArr v_b^2 = v_(b,r)^2 - v_r^2` `rArr v_b^2 = sqrt(v_(b,r)^2 - v_r^2) = sqrt(4^2 - 2^2) = 2 sqrt(3) m s^-1` ( c) Time taken to cross the river is (Width of river)/(Velocity of boat w.r.t Earth) `rArr t = (800)/(2 sqrt(3)) = (400 sqrt(3))/(3) s`.  .
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| 14180. |
Four identical particles each of mass 1kg are arranged at the corners of a square of side length 2sqrt(2)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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| 14181. |
Define relative velocity. |
| Answer» Solution :When two objects A and B are MOVING with different VELOCITIES , then the velocity of one OBJECT A with respect to another object B is called relative velocity of object A with respect to B . | |
| 14182. |
Three metal rods made of copper, brass and steel each of area of cross-section 4 cm^(2) are joined as shown in the figure. Their lengths are respectively 46 cm, 13 cm and 12 cm. Their coefficients of thermal conductivity are respectively 0.92, 0.26 and 0.12, all in CGS units. The rods are thermally insulated from the surroundings except at the ends. Rate of flow of heat through the copper rod. in cal S^(-1) is |
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Answer» 2.4 |
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| 14183. |
At 27^(@)C , two moles of an ideal mono-atomic gas occupy a volumeV. The gas expands adiaba-tically to a volume 2V. Calculate (a) final temperature of the gas (b) Change in its internal energy and (c ) the work done by the gas during the process. |
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Answer» Solution :(a) In CASE of ADIABATIC change, `PV^(Gamma)` = const. with PV=nRT So that `T_(1)V_(1)^(gamma-1) = T_(2)V_(2)^(gamma-1) ["with " gamma = (5//3)]` i.e., `300xxV^(2//3) = T(2V)^(2//3) or T=300//(2)^(2//3)=189K` (b) As `Delta U=(2xx8.31xx(189-300))/([5/3-1]) = -2767.23 J` NEGATIVE sign means internal energy decreases. (c) According to first law of thermodynamics, `DeltaQ=DeltaU+DeltaW` and as for adiabatic change `DeltaQ=0 rArr DeltaW = -DeltaU=2767.23J` . |
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| 14184. |
Two satellite A and B, ratio of masses 3:1 are in circular orbits of radii r and 4r. Then ratio mechanical energy of A and B is |
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Answer» `1:3` `E=-(GMm)/(2r)or E prop (m)/(r)` `rArr (E_(A))/(E_(B))=(m_(A))/(m_(B))XX(r_(B))/(r_(A))=(3)/(1)xx(4r)/(r)""{because (m_(A))/(m_(B))=(3)/(1),(r_(B))/(r_(A))=(4r)/(r)}` `rArr (E_(A))/(E_(B))=(12)/(1)`. |
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| 14185. |
What is the effect of temperature on elasticity? |
| Answer» SOLUTION :As TEMPERATURE of the substances increases, its ELASTICITY DECREASES. | |
| 14186. |
(A) To increase the sensitivity of a camera, its aperture should be reduced. ( R) Smaller the aperture, image focussing is also sharp. |
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Answer» Both A and R are TRUE and R is the CORRECT EXPLANATION of A |
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| 14187. |
A liquid of densilty 1200 kg//m^(3) flowing steadily in a tube of varying cross section. The cross section at a ponit A is 1.0cm^(2) and that at B is 20mm^(2), the ponts A and B are in the same horizontal plane. The speed of the liquid at A is 10cm/s. Calculate the difference i pressure at A and B. |
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Answer» Solution :From equationof continuity, the speed `v_(2)` at B is given by `A_(1)v_(1)=A_(2)v_(2)` or `(1.0)^(2)(10)=(20xx10^(-2))v_(2)` or `v_(2)=1.0/(20xx10^(-2))xx10=50`cm/s Ber Bernoull.s equation `P_(1)+rhogh_(1)+1/2rhov_(1)^(2)+P_(2)+RHO gh_(2)+1/2rhov_(2)^(2)` Here `h_(1)=h_(2)` Thus `P_(1)-P_(2)=1/2 rho v_(2)^(2)-1/2rhov_(1)^(2)` br> `=12XX(1200)(2500-100)` `=600xx2400=144` Pa |
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| 14188. |
Four particles each of mass .m. are located at the four corners of a square of side a. If the system is revolving in a circle due to their mutual force of attraction then the angular velocity and time period of each particle is |
Answer» Solution : The gravitational force between the masses provides the necessary centripetal force `(Gm_1m_2)/(d^2)=m_(1)r_(1)omega^2 RARR(1)` The distance of CENTRE of mass from `m_1` is `r_(1)=(m_2d)/(m_1+m_2)rarr(2)` From (1) and (2) `(Gm_1m_2)/d^2 =(m_1m_2d)/(m_1+m_2).omega^2` (or) `omega^(2)=(G(m_1+m_2))/d^3 ` (or) `omega=sqrt((G(m_1+m_2))/d^3)` The resultant gravitational force on one of the particles GM^2)/a^2+(Gm^2)/(2a^2)` This gravitational force provides the necessary centripetal force `mromega^2=(Gm)/(a^2)(SQRT2+1/2)implies` `a/sqrt2omega^2=(Gm)/a^2((2sqrt2+1)/2)impliesomega=sqrt((Gm)/a^3(2+1/sqrt2)) and T =(2pi)/omega=2pisqrt((a^3)/(Gm)((sqrt2)/(2sqrt2+1))` |
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| 14189. |
How much work is done by agent in forcing 3m^(3) of water through a pipe of radius 2cm, if the difference in pressure at the two ends of the pipe is 10 rhoa? |
| Answer» SOLUTION :`3XX10^(6)J` | |
| 14190. |
A spring is compressed between two toy cars. When the cars are released, they move apart. If x_1and x_2are displacements of the cars when in contact with the spring, then (mass ae m_1 and m_2) |
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Answer» `m_1x_1 = m_2x_1` |
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| 14191. |
Consider a horse attached to the cart which is initially at rest. If the horse starts walking forward, the cart also accelerates in the forward direction. If the horse pulls the cart with force F_hin forward direction, then according to Newton's third law, the cart also pulls the horse by equivalent opposite force F_c = F_hin backward direction. Then total force on cart+horse' is zero. Why is it then the 'cart+horse'accelerates and moves forward? |
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Answer» Solution :This paradox arises due to wrong application of Newton.s second and THIRD laws. Before applying Newton.s laws, we should decide .what is the system?.. Once we identify the .system., then it is possible to identify all the forces acting on the system. We should not consider the force exerted by the system. If there is an UNBALANCED force acting on the system, then it should have acceleration in the DIRECTION of the resultant force. By following these steps we will analyse the horse and cart motion. If we decide on the cart+horse as a .system., then we should not consider the force exerted by the horse on the cart or the force exerted by cart on the horse. Both are internal forces acting on each other. According to Newton.s third law, total internal force acting on the system is zero and it cannot accelerate the system. The acceleration of the system is caused by some external force. In this case, the force exerted by the road on the system is the external force acting on the system. It is wrong to conclude that the total force acting on the system (cart+horse) is zero without including all the forces acting on the system. The road is PUSHING the horse and cart forward with acceleration. As there is an external force acting on the system, Newton.s second law has to be applied and not Newton.s third law. The following figures illustrates this. If we consider the horse as the . system ., then there are three forces acting on the horse. (i) Downward gravitational force `(m_h g)` (ii) Force exerted by the road `(F_r)`(iii) Backward force exerted by the cart `(F_c)` It is shown in the figure. ` F_r`- Force exerted by the road on the horse `F_c ` - Force exerted by the cart on the horse `F_e ` -Perpendicular component of `F_r = N` ` F_r ` Parallel component of `F_r`which is reason for forward movement  The force exerted by the road can be resolved into parallel and perpendicular components. The perpendicular component balances the downward gravitational force. There is parallel component along the forward direction. It is greater than the backward force `(F_c)` . So there is net force along the forward direction which causes the forward movement of the horse. If we take the cart as the system, then there are three forces acting on the cart. (i) Downward gravitational force `(m_c g)` (ii) Force exerted by the road `(F_r)` (iii) Force exerted by the horse` (F_h)` It is shown in the figure  The force exerted by the road `(vecF_r )`can be resolved into parallel and perpendicular components. The perpendicular component cancels the downward gravity `(m_c g)` . Parallel component acts backwards and the force exerted by the horse `(vecF_h)`acts forward. Force `(vecF_h)`is greater than the parallel component acting in the opposite direction. So there is an overall unbalanced force in the forward direction which causes the cart to accelerate forward. If we take the cart+horse as a system, then there are two forces acting on the system. (i) Downward gravitational force `(m_h + m_c)` (ii) The force exerted by the road `(F_r)`on the system. It is shown in the following figure.  (iii) In this case the force exerted by the road `(F_r)`on the system (cart+horse) is resolved in to parallel and perpendicular components. The perpendicularcomponent is the normal force which cancels the forward movement of downward gravitational force `(m_h + m_c)g` . The parallel component of the force is not balanced, hence the system (cart+horse) accelerates and moves forward due to this force. |
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| 14192. |
The lengths of two bodies measured by a metre scale are l_1=(20_-^+0.5)cmand l_2=(15_-^+0.2)cm Calculate: Difference between the lengths. |
| Answer» SOLUTION :`(5_-^+0.7)CM` | |
| 14193. |
A springof spring constant 500 N/m is attached on a rough surface at one side . Coefficient of friction for rough surface is 0.75. A block of mass 100 kg collide with the free end of spring at a speed of 10sqrt(2) ms^(-1), then how much will spring be compressed ? (g =10 ms^(-2)) |
| Answer» SOLUTION :X = 5M | |
| 14194. |
The density of a liquid in CGS system is 0.625g//cm^(3). What is its magnitude in SI system? |
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Answer» `625kg//m^(3)` `1g=10^(-3)kg` `1cm=10^(-2)m` `therefore"Density"=(0.625xx10^(-3))/((10^(-2))^(3))` `=0.625xx10^^(-3)xx10^(6)` `=0.625xx10^(3)=625kg//m^(3)` |
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| 14195. |
An iron needle slowly placed on the surface of water floats on it because |
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Answer» when inside in water it will displace water more than its WEIGHT |
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| 14196. |
A cubical copper block has eahc side 2.0 cm. It is suspended by a string and submerged in oil of density 820 kg m^(-3). Calculate the tension in the string. (density of copper = 8920 kg m^(-3)) |
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Answer» Solution :Tension = true WEIGHT - upthrust DUE to oil `= V rho g - V sigma g = V (rho - sigma) g` `=(2 xx 10^(-2))^(3) xx (8920 - 820)xx 10 =8 xx 10^(-6) xx 8100 xx 10`. |
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| 14197. |
1/2 mole of Helium gas is contained in a container at S.T.P. The heat energy needed in cal to doublethe pressure of the gas, keeping the volume constant |
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Answer» 3276 J |
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| 14198. |
If the weight of body is greater than the buoyant force of liquid then this body float on the surface of liquid. |
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Answer» |
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| 14199. |
In the case of motion of a fluid in a tube where area of cross section is maximum a) velocity is maximum b) pressure is maximum c) velocity is minimum d) pressure is minimum |
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Answer» B, C are correct |
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| 14200. |
A pipe open at both ends produces a note of frequency f_(1) . When the pipe is kept with (4)/(3)th of its length it water, it produced a note of frequency f_(2). The ratio (f_(1))/(f_(2)) is |
| Answer» Answer :C | |