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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.
| 14901. |
Two water pipes of diameter 2cm and 4 cm are connected to a water tank at the same depth. The length of 4 cm pipe is four times that of the length of 2 cm pipe. The rate of water flow, through 2 cm pipe as compared to that through 4 cm pipe will be |
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Answer» ONE fourth |
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| 14902. |
(A) : Graph between pressure P and h below the surface of a liquid open to atmosphere is shown |
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Answer» Both (A) and (R ) are TRUE and (R ) is the correct EXPLANATION of (A) |
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| 14903. |
Three vector vec(A)=ahat(i)+hat(j)+hat(k) , vec(B)=hat(i)+bhat(j)+hat(k)and vec(C)=hat(i)+hat(j)+chat(k) aremutually perpendicular (hati , hatj and hatk) unit vectors along X,Y and Z axis respectively).The respective values of a,b and c are |
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Answer» 0,0,0 |
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| 14904. |
A tap supplies water at 10^@Cand another tap at 100^@C. How much hot water must be taken so that we get 20kg of water at 35^@C . |
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Answer» 40/9 KG |
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| 14905. |
Two circular loops of radii R and nR are made of same wire. If their M.I. about their normal axis through centre are in the ratio 1:8 the value of n is. . . . . . . . . |
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Answer» |
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| 14906. |
At a given temperature,the ratio of rms velocity of hydrogen to rms velocity of oxygen is |
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Answer» 4 |
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| 14907. |
Derive an expression for the moment of inertia of a thin spherical shell about a diameter. |
Answer» Solution :Consider a solid sphere. Suppose a smaller concentric sphere is removed from the solid sphere we get a hollow or a thick spherical shell (See Fig. 7.2.58). Let Mbe the mass, R the external radius and r the internal radius of the hollow sphere. Volume of the shell `=4/3 pi(R^(3)-r^(3))`  Mass PER unit volume of the shell `=rho` `rho = M/((4/3)pi(R^(3)-r^(3)))` `=(3M)/(4PI(R^(3)-r^(3))` M.I. of the solid sphere of radius R about a diameter `=2/3 XX "mass" xx ("radius")^(2)` `=2/5 xx 4/3pi R^(3) rho R^(2)` `=8/15 pi R^(5) rho` M.I. of the solid sphere of radius r about a diameter `=8/15 pi r^(5) rho` M.I. of the hollow sphere about the diameter = `L` `=8/15 piR^(5) - 8/15 pi r^(5) rho = 8/15 pi rho (R^(5)-r^(5))` `l=2/5 M (R^(5)-r^(5))/(R^(3)-r^(3))` |
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| 14908. |
From a complete ring of mass M and radius R, an are of making 30^(@) at centre is removed. What is the moment of inertia of the incomplete ring about an axis passing through the centre of the ring and _|_^(et) to the plane of the ring . |
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Answer» Solution :Mass of in COMPLETE ring`= M -(M)/(2pi)xx(pi)/(16)=(11M)/(12)` MOI of INERTIA of INCOMPLETE ring `I=[(11m)/(12)]R^(2)` `I=(11)/(12)MR^(2)` 
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| 14909. |
Due to the action of internal forces of the system, the total linear momentum of the system is |
| Answer» Solution : always zero | |
| 14910. |
At 0 k, which of the following properties of a gas is zero? |
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Answer» KINETIC energy |
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| 14911. |
Calculate the gravitational potential near the surface of the Earth. |
| Answer» Solution :The GRAVITATIONAL POTENTIAL at this point is SIMPLY `V(h)=U(h)//m=gh`. | |
| 14912. |
Name the method of heat transfer which does not depends on gravity ? |
| Answer» SOLUTION :RADIATION and CONDUCTION | |
| 14913. |
Explain why, an optical pyrometer (for measuring high temperatures) calibrated for an ideal black body radiation gives too low a value for the temperature of a red hot iron piece in the open, but gives a correct value for the temperature when the same piece is in the temperature when the same piece is in the furnace. |
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Answer» Solution :Optical pyrometer is developed on the principle of lightening of surface depends on TEMPERATURE. When the temperature is less than `600^(@)C` (iron piece is in open) then it is not bright, HENCE temperature in optical pyrometer is low. But when that piece is kept in FURNACE, then it is bright and optical pyrometer gives correct value of temperature. |
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| 14914. |
An open pipe is suddenly closed at one end with the result that the frequency of the third harmonic of the closed pipe is found to be higher by 100 Hz than the fundamental frequency of the open pipe. What is the fundamental frequency of the open pipe? |
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Answer» Solution :Fundamental frequency of open pipe, `v_(o)=(v)/(2L)` Frequency of THIRD harmonic of CLOSED pipe, `v_(C)=(3v)/(4L)` `:.(v_(C))/(v_(O))=(3)/(2)` (or) `v_(C)=(3)/(2)v_(O)` Given `v_(C)=v_(O)+100` `:.(3)/(2)v_(O)=v_(O)+100` Fundamental frequency (or) `v_(O)=200Hz`. |
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| 14915. |
A 16cm^3 volume of water folws per second through a capillary tube of radius r cm and of length 1 cm, when connected to a pressure head of h cm of water. If a tube of the same length and radius r/2 is connected to be same pressure head, find the mass of water flowing per minute through the tube. |
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Answer» Solution :Here, `V_1 = 16cm^3//"sec" , P_1 = h rho g, r_1 = r , l_1 = l , r_2 = r//2 , P_2 = h rho g`, So `P_1 = P_2 ,` Now `V_1 = (PI P_1 r_1^4)/(8eta l_1) and V_2 = (pi P_2 r_2^4)/(8 eta l_2) , (V_2)/(V_1) = (P_2)/(P_1) XX (r_2^4)/(r_1^4) xx (l_1)/(l_2) = ((r//2)^4)/(r^4) xx l/l = (1/2)^4 = 1/(16)` `V_2 = (16)/(16)= 1 cm^3//s` , volume of water flowing PER minute `=1 xx 60 = 60 cm^3//"MIN"` `therefore ` Mass of water flowing per minute `= 60 xx 1 = 60 ` gram/min |
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| 14916. |
Given is the graph between PV/T and P for 1 gm of oxygen gas at two different temperature T_(1) and T_(2) (Given:Density of oxygen=1.429 kg//m^(3)).The value of PV/T at the point A and the relation between T_(1) and T_(2) and respectively |
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Answer» 0.256 `JK^(-1)` and `T_(1)ltT_(2)` |
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| 14917. |
A thin equiconvex lens is made of = glass of refractive index 1.5 and its focal length in air is 0.2m. If it acts as a concave lens of 0.5m focal length when dipped in a liquid, the velocity of light in the liquid is |
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Answer» `1.2 TIMES 10^8 ms^-1` |
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| 14918. |
Gravitational force between two point masses m and M separated by a distance r is F. Now if a point mass 3m is placed next to m, what will be the (a) Force on M dur to m (b) total force on M ? |
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Answer» Solution :The Gravitational FORCE on m due to M `F = (GMm)/r^2` a)Since the gravitational force between two point MASSES is independent of the presence of other masses , So if a point mass 3m is placed next to m, the force doesn.t CHANGE .`"" :. F = (GMm)/r^2` b) Total force on body of mass M is `F_(1)=(Gm(3m+m))/r^2 = (4GMm)/r^2=4F` |
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| 14919. |
What is the relation between loudness and intensity? |
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Answer» Solution :According to Weber-Fechner's law, loudness (L) is PROPORTIONAL to the logrithm of the actual intensity (I) MEASURED with an accurate non-human INSTRUEMENT. '' This means that `L prop LN I` |
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| 14920. |
Force vec(F ) = (60 hat(i) +50 hat(j)-3hat(k))Nis applied on a particle , the value of velocity is vec(v) =(2hat(i)+4hat(j)+5hat(k)) m/s , then power will be ……. W. |
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Answer» 305 `= 120 - 60 - 15` = 45 W . |
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| 14921. |
A bob of mass m is suspended by a light string of length L. It is imparted a horizontal velocity v_(o) at the lowest point A such that it completes a semi-circular trajectory in the vertical plane with the string becoming slack only on reaching the topmost point, C. This is shown in Fig. 6.6. Obtain an expression for v_(o) |
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Answer» SOLUTION :There are two external FORCES on the bob : gravity and the tension (T) in the string. The latter does no work since the displacement of the bob is always normal to the string. The POTENTIAL energy of the bob is thus associated with the gravitational force only. The total mechanical energy E of the SYSTEM is conserved. We take the potential energy of the system to be zero at the lowest point A. Thus, at A , `E= (1)/(2)mv_(0)^(2)""(6.12)` `T_(A)-mg= (mv_(0)^(2))/(L)` [Newton.s Second LAW] where `T_(A)` is the tension in the string at A. At the highest point C, the string slackens, as the tension in the string `(T_C)` becomes zero. Thus, at C `E= (1)/(2)mv_(c )^(2)+ 2mgL""(6.13)` `mg= (mv_(c )^2)/(L)` [Newton.s Second Law] (6.14) where `v_(c )` is the speed at C. From Eqs. (6.13) and (6.14) `E= (5)/(2)mgL` Equatting this to the energy at A `(5)/(2)mgL= (m)/(2)v_(0)^(2)` or, `v_(0)= sqrt(5gL)`. |
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| 14922. |
A piece of charcoal and a piece of shining steel of the same area are kept for a long time in an open lawn in bright sun. a) The steel will absorb more heat than the charcoal b) The temperature of the steel will be higher than that of the charcoal c) If both are picked up by bare hands, the steel will be left hotter than the charcoal d) If the two are picked up from the lawn and kept in a cold chamber, the choreal will lose heat at a faster rate than the steel |
| Answer» ANSWER :C | |
| 14923. |
A bob of mass m is suspended by a light string of length L. It is imparted a horizontal velocity v_(o) at the lowest point A such that it completes a semi-circular trajectory in the vertical plane with the string becoming slack only on reaching the topmost point, C. This is shown in Fig. 6.6. Obtain an expression for The speeds at points B and C. |
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Answer» Solution :It is clear from Eq. (6.14) `v_(c )= sqrt(gL)` At B, the energy is `E=(1)/(2)mv_(B)^(2)+mgL` Equatting this to the energy at A and employing the RESULT from (i), namely `v_(0)^(2)= 5gL`, `(1)/(2)mv_(B)^(2)+mgL= (1)/(2)mv_(0)^(2)` `(5)/(2)mgL` `v_(B)= sqrt(3gL)`. |
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| 14924. |
Three discs, A, B and C having radii 2 m, 4 m and 6 m respectively are coated with carbon black on their outer surfaces. The wavelengths corresponding to maximum intensity are 300 nm, 400 nm and 500 nm, respectively. The power radiated by them are Q_(A) , Q_(B) and Q_(c) respectively. |
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Answer» `Q_(A) ` is MAXIMUM |
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| 14925. |
A thin metal disc of radius of 0.25 m and mass 2 kg starts from rest and rolls down on an inclined plane. If its rotational kinetic energy is 4 J at the foot of inclined plane, then the linear velocity in m/s) at the same point is |
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Answer» 2 Here , ` I = (MR^(2))/(2) , THEREFORE (1)/(2) (MR^(2))/(2) omega^(2) = 4 J` or `(1)/(2) X (2)/(2) (R omega)^(2) = 4 J` Since V = `R omega , therefore v^(2) = 8 or v = 2 sqrt2` m/s |
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| 14926. |
A bob of mass m is suspended by a light string of length L. It is imparted a horizontal velocity v_(o) at the lowest point A such that it completes a semi-circular trajectory in the vertical plane with the string becoming slack only on reaching the topmost point, C. This is shown in Fig. 6.6. Obtain an expression for The ratio of the kinetic energies (K_(B)"/"K_(C )) at B and C. Comment on the nature of the trajectory of the bob after it reaches the point C. |
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Answer» Solution :The ratio of the kinetic energies at B and C is : `(K_B)/(K_C)=((1)/(2)mv_(B)^(2))/((1)/(2)mv_(c )^(2))=(3)/(1)` At point C, the string becomes slack and the velocity of the BOB is horizontal and to the left. It the connecting string is cut at this instant, the bob will execute a projectile motion with horizontal projection akin to a rock kicked HORIZONTALLY from the edge of a cliff. OTHERWISE the bob will continue on its circular PATH and complete the revolution. |
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| 14927. |
What is an echo? what is the minimum distance required for echo? |
| Answer» Solution :ECHO is the phenomenon of REPETITION of sound due to its reflection from the SURFACE of a large OBSTACLE. | |
| 14928. |
The amplitude of a simple pendulum decreases with time. But there is no decrease in the amplitude of a pendulum clock. Why? |
| Answer» SOLUTION :In the case of simple pendulum amplitude DECREASES because energy is LOST in OVERCOMING AIR resistance, friction etc. In the case of pendulum clock this energy lost is balanced by the energy supplied by the wound spring. | |
| 14929. |
A system is taken from state A to state B alongtwo different paths 1 and 2. The heat absorbed and work done by the system along these two paths are Q_(1) and Q_(2), and W_(1) and W_(2)respectively. |
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Answer» `Q_(1)=Q_(2)` |
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| 14930. |
The coefficient of static friction between contact surfaces of two bodies is 1. The contact surfaces of one body support the other till the inclination is less than |
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Answer» `30^(@)` |
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| 14931. |
A body is hanging from a rigid support by an inextensible string of length l. it is struck inelastically by an identical body of mass m with horizontal velocity v = sqrt(2gl), the tension in the string increases just after striking by |
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Answer» mg |
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| 14932. |
A wave of frequency 500 Hz has a velocity 300 m/s. The distance between two nearest points which are 60^(@) out of phase , is ………… . |
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Answer» 0.2 m `LAMBDA = (300)/(500) = 0.6` m `Delta X = (lambda)/(2pi) * Delta phi = (0.6)/(2pi) * (pi)/(3) = 0.1 ` m |
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| 14933. |
Explain principle of super position forgravitational force. |
Answer» Solution :`implies` The net gravitational force on a particle is the vector sum of the INDIVIDUAL gravitational FORCES on the particle.  `implies` Gravitational force on point mass `m_1` due to point mass `m_2` `vecF_(12)=G(m_1m_2)/(|vecr_(12)|).hatr^2""`or `vecF_(12)=(-Gm_(1)m_(2))/(|vecr_(21)|)^2.hatr_(21) ""...(1)` `hatr_(12)` is a UNIT vector from `m_(1) "to " m_2` `hatr_(21)` is a unit vector from `m_(2) "to " m_1` `implies` Gravitational force of mass `m_1` due to mass `m_3` `vecF_(13)=(Gm_1m_3)/(|vecr_(13)|).hatr_(13)=(-Gm_(1)m_(2))/(|vecr_(31)|^2)hatr_(31)""...(2)` `implies` Gravitational force of mass `m_1` due to mass `m_4` `vecF_(14)=(Gm_1m_4)/(|vecr_(14)|).hatr_(14)=(-Gm_(1)m_(4))/(|vecr_(41)|^2)hatr_(41)""...(3)` `implies` The TOTAL force on `m_1` `vecF_(1)=(Gm_1m_2)/(|vecr_(12)|).hatr_(12)+G(m_(1)m_(2))/(|vecr_(13)|^2)hatr_(31)+(Gm_(1)m_(4))/(|vecr_(14)|^2)""...(4)` `vecF_(1)=-((Gm_1m_2)/(|vecr_(21)|).hatr_(21)+G(m_(3)m_(1))/(|vecr_(31)|^2)hatr_(31)+(Gm_(4)m_(1))/(|vecr_(41)|^2)hatr_(41))""..(5)` Note : The gravitational field intensity is the force on a unit mass at a point in the field. |
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| 14934. |
Three particle each of 1kg mass are placed at corners of a right angled triangle AOB, .O. being the origin of coordinate system (OA and OB along the x - direction and + ve y-direction). If OA = OB = 1m, the position vector of centre of mass (in metres) is |
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Answer» `(hati+hatj)/(3)` |
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| 14935. |
A geyser heats water flowing at the rate of 3 kg per minute from 27^(@)C to 77^(@)C . If the geyser operates on a gas burner, what is the rate of consumption of fuel if the heat of combustion is 4 xx 10^(4) J//g? Given specific heat of water is 4.2 xx 10^(3) J//kg//K. |
| Answer» Solution :Mass of fuel rewuired `= 1 G XX 4200 xx 50//4 xx 10 ^(4) = 15.75g` PER minute | |
| 14936. |
The surfaces of the train wheels and railway tracks are made smooth ". Why ? |
| Answer» SOLUTION :So that they are in good CONTACT with each other, DUE to increased VALUE of friction. | |
| 14937. |
Choose the correct statement. (a) Any motion that repeats itself in equal intervals of time along the same path is called periodic motion. (b) The displacement of a particle in periodic motion can always be expressed in terms of sine and cosine functions of time. (c) A body in periodic motion moves back and forth over the same pah is called oscillatory or vibrating motion (d) Simple harmonic motion is a particular case of periodic motion. |
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Answer» Only a, B, d are TRUE |
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| 14938. |
A vehicle whose wheel track is 1.7m wide and whose centre of gravity 1 m above the road and central between the wheels, taken a curve whose radius 50m, on a level road. The speed at which the inner wheel would leave the road is close to n xx 5m//s. |
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Answer» |
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| 14939. |
An aeroplane travelling towards north at 200 ms^-1turns east and travel at the same rate. Rs change in velocity is |
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Answer» `100ms^-1` |
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| 14940. |
A train of weight 10^7 N is running on a level track with uniform speed of 36 km h^(-1). The frictional force is 0.5 kg f per quintal. If g = 10 ms^(-2), power of engine is |
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Answer» Solution :Weight of train = `10^(7) N = 10^6 kg f` Frictional force, `f = (0.5)/(100) xx 10^(6) = 5000 kg, N= 50000 N` `v = 36 km H^(-1) = 10 ms^(-1)` The power of engine is `P = f xx v = 50000 xx 10 = 5 xx 10^(5) W = 500 k W ` . |
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| 14941. |
A mass M attached to a horixontal spring executes SHM with an amplitude A_(1) . When mass M passes through its mean position a smaller mass m is placed over it and both of them move together with amplitude A_(2). The ratio of ((A_(1))/(A_(2))) is |
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Answer» Solution :`T_(1)= 2pisqrt((M)/(K)), T_(2)= 2pisqrt(((m+M))/(k))` Using law of conservation of linear MOMENTUM `MV_(1)= (m+M)V_(2)` `M(A_(1)omega_(1))= (m+M)(A_(2)omega_(2))` `(A_(1))/(A_(2))= ((m+M))/(M)(omega_(2))/(omega_(1))= ((m+M))/(M)sqrt((k)/(m+M)).sqrt((M)/(k))= sqrt(((m+M))/(M))` |
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| 14942. |
The angle which the rough inclined plane makes with the horizontal when the body placed on it just starts sliding down is called |
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Answer» angle of FRICTION |
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| 14943. |
(A) : The coefficient of restitution for a perfectly elastic collision is equal to one.(R ) :In perfectly elastic collisio, kinetic energy before and after the collision remains conserved. |
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Answer» Both (A) and (R ) are true and (R ) is the correct EXPLANATION of (A) |
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| 14944. |
For a monoatomic gas at temperature T, match the following. |
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Answer» <P> |
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| 14945. |
A block of mass is moving with a constant acceleration on a rough plane. If the coefficient of friction between the block and the ground is mu, the power delivered by the external agent after a time t from the beginning is equal to |
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Answer» `ma^(2)t` |
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| 14946. |
A wire suspended vertically from one end is stressed by attaching a weight of 200N to lower end the weight stretched the wire by 1 mm. the elastic energy stored the wire is |
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Answer» 20J |
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| 14947. |
A one litre flask contains some mercury. It is found that at different temperature the volume of air inside the flask remains the same. What is the volume of the mercury in the flask? (alpha of glass is 9 xx 10 ^(-6)//^(@)C and Y _(R) of Hg is 1.8 xx 10 ^(-4) //^(@)C) |
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Answer» Solution :Coefficient of volume expansion of the flask `(gamma) = 3 alpha = 3 xx 9 xx 10 ^(-6) = 27 xx 10&(-6) //^(@)C` Coefficient of volume expansion of `Hg = 1.8 xx 10 ^(-4) //^(@)C` Total volume of the flask (v) = 1 LITRE `= 1000 cm ^(3).` Let .x. be the volume of the mercury in the flask. If the expansion of teh mercury is same as the expansion of the flask then the space in the flask remains the same. Expansion of te flask for the rise of `t ^(@)C = v gamma t` Expansion of the mercury for the rise of `t ^(@)C = v gamma _(g) t therefore x gamma _(R )t = v gamma t` `implies x = (v gamma)/( gamma _(R)) = 100 xx (27 xx 10 ^(-6))/( 1.8 xx 10^(-4)) = 150 cm ^(3)` |
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| 14948. |
A block of mass m is suspended by a light thread from an elevator. The elevator starting from rest is accelerating upward with uniform acceleratio a. The work done by tension on the block during t seconds is |
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Answer» `m/4[g+a]at^(2)` |
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| 14949. |
In the pulley system shown, if radii of the bigger and smaller pulley are 2 m and 1m respectively and the acceleration of block A is "5m/s"^(2) in the downward direction, then the acceleration of block B will be : [Pulleys are fixed to each other] |
| Answer» Answer :D | |
| 14950. |
Calculate the difference between the two specific heats of nitroge, given that the density of nitrogen at N.T.P is 1.25 g/litre and J = 4200 J/kcal. Express it in kcal/kgK. |
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Answer» Solution :The differncein SPECIFIC heats, `r=(P)/(rho T)`. `P= 1.013xx10^(5)N//m^(2), T=273K`, `rho=` density of the gas `rho = 1.25 XX (10^(-3) kg)/( 10^(3)cm^(3)) =(1.25 xx 10^(-3) kg)/( 10^(3) xx 10^(-6) m^(3))= 1.25 kg//m^(3)` `:.r=(1.013 xx 10^(5))/(1.25 xx 273) = 296.8J//kgK` `r=(296.8)/( 4200) "kcal"//kgK=0.07068`kcal/kg K. `:.` The difference of specific heats = 0.0768 kcal/kg K. |
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