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14351.

A body is sliding down an inclined plane have coefficient of friction 0.5. If he normal reaction is 4 times that of resultant downward of force along the incline, find the angle between the inclined plane and the horizontal.

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`45^(@)`
`53^(@)`
`37^(@)`
`60^(@)`

ANSWER :C
14352.

The density of mercury in cgs system is 13.6 g cm^(-3). Its value in SI is

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`136 kg//m^(3) `
`1360 kg//m^3`
`13600 kg//m^3`
`1.36 kg//m^3`

ANSWER :C
14353.

List the points to be considered while calculating the torques on rigid bodies.

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SOLUTION :(i) Only those FORCES that lie on planes PERPENDICULAR to theaxis and
(ii) POSITION VECTORS which are perpendicular to the axis.
14354.

The radius of earth is 6.37xx10^6m, its mean density 5.5xx10^3 kgm^-3 and G = 6.66xx10^-(11) Nm^2 kg^-2. Determine the gravitational potential on the surface of earth.

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zero
2
infinity
depends on MASS of bob

Answer :C
14355.

Assertion: For an ideal gas, at constant temperature, the product of the pressure and volume is constant. Reason:The mean square velocity of the molecules is inversely proportional to mass.

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Both ASSERTION and reason are true and reason is the correct EXPLANATION of the assertion.
Both assertion and reason are true but reason is not the correct explanation of the assertion .
Both assertion and reason are false
Assertion is true but reason is false

Solution :According to Boyle's low PV - constant at a GIVEN TEMPERATURE.
14356.

A lead ball is allowed to fall through an elongated column filled with glycerine. What sort of graph would we get on plotting the velocity (v) and the distance traversed (s) by the lead ball?

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SOLUTION :Glycerine is a viscous liquid. We know that the velocity of the lead ball will increase at first, and, after sometime, it will move with a uniform velocity (terminalvelocity). The v-s GRAPH or velocity-displacement graph OBTAINED is as shown in Fig. 3.39.
14357.

A 500kg boat has an initial speed of 10 ms^(-1)as it passes under a bridge. At that instant a 50 kg man jumps straight down into the boat from the bridge. The speed of the boat after the man and boat attain a common speed is

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`100/11 MS^(-1)`
`10/11 ms^(-1)`
`50/11 ms^(-1)`
`5/11 ms^(-1)`

ANSWER :A
14358.

A body is pojected with a velocity of 30m/s to have a horizontal range of 45m. Find the angle of projection.

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ANSWER :`15^(@)` and `75^(@)`
14359.

Give the names of physical quantities which are needed to describe fluid.

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SOLUTION :Pressureand DENSITY.
14360.

The unit vector along the negative z-axis is :

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`-HAT(K)`
`-hat(J)`
`hat(i)`
`+hat(k)`

ANSWER :A
14361.

Galileo writes that for angles of projection of a projectile at angles (45^(@)+theta) and (45^(@)-theta), the horizontal ranges described by the projectile are in the ratio of (ifthetale45^(@))

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`2:1`
`1:2`
`1:1`
`2:3`

ANSWER :C
14362.

Define periodic time and angular frequency and obtain the relation between them.

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Solution :Periodic time : ..The time required to COMPLETE one rotation is called periodic time (T)...
Angular frequency : `2pi` time the frequency of an oscillator is called angular frequency `(omega)`
`therefore omega = (2pi)/(T)`
The displacement of SHM PARTICLE with amplitude A and INITIAL phase `phi=0` at time t is `x(t)= A sin omega t"""........."(1)`
Since the MOTION has a period T, hence
`x(t)= x(t+T)" That is "A COS omega t= A cos (omega t + phi)"""........."(2)`
Now the cosine function is periodic with period `2pi` means phase increase by `2pi` and it motion repeared.
`therefore omega t+2pi = omega(t+T)`
`therefore omega t+ 2pi = omega t+ omega T`
`therefore 2pi = omegaT`
`therefore omega = (2pi)/(T)`
but `(1)/(T)=v` (frequency)
`therefore omega= 2pi v`
So `pi" is "2pi` time the frequency of oscillation `v(1/T)`.
14363.

A ball is released from a height of 10m. If it loses 20% of its energy on hitting the ground, the height to which it bounces is

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7.5
8m
9m
6m

Answer :B
14364.

Is the specific heat of water and ice be the same? Give their values.

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Solution :No, specific HEAT of water is ` 1 "G CM"^(-3)` and specific heat of ice is `0.5 "g cm"^(-3)`
14365.

A clean body of mass 100g starts with a velocity of 2 m//s on a smooth horizontal plane , accumulating dust at the rate of 5 g//s . Find the velocity at the end of 20 seconds and the distance travelled during that period .

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ANSWER :`27.7m`
14366.

A body of mass 1 kg falls from rest through a distance of 200 m and acquires a speed of 50 m/s. Work done against friction of air is (Take g = 10 m//s^2)

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700 J 
1250 J 
750 J 
1960 J 

SOLUTION :PE lost = `mgh = 1 xx 10 xx 200 = 2000 J`
KE gained = `1/2 mv^2 = 1/2 xx 1 xx (50)^2 = 1250 J`
Work done against friction of AIR = `2000 J - 1250 J = 750 J` .
14367.

A block of mass 15 kg is lying on an inclined plane of angle 30^(@). In order to make it move upward along the slope with an acceleration 25 cms^(2), a horizontal force of 200 N is required to be applied on it. Then the frictional force on the block i s (g = 9.8 ms 2)

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95.95 n
134 n
99.70 n
90 n

Solution :
Fromthe figureif theblockis movingupwardalongthe slopewith ANACCELERATION`a_(x )`them `Sigma F_(x ) ne0 ` but `Sigma F_(x )= ma_(x )`
`200 xx0.866-f -15 xx 9.8 xx (1)/(2) =15 xx 0 .25`
`173.2 -f -73.5 = 3.75`
`:. ,95 . 95 N=f`
14368.

Define force constant of spring .

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SOLUTION :The spring CONSTANT of a spring is the change in the force it exerts, DIVIDED by the change in deflection of the spring. (K = f/x)
14369.

Irreversible process is

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ADIABATIC process
Joule-Thomson expansion
Ideal ISOTHERMAL process
None of the above

Answer :B
14370.

A string wound aroung a solid cylinder is unwound by applying a force of 20 N. Calculate the torque acting on the cylinder and angular acceleration produced in the cylinder of mass 2.5 kg and of radius 50 cm.

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ANSWER :10 N.m; 32 RAD .`s^(-1)`
14371.

What are the differences between connection and conduction?

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Solution :Conduction: Conduction is the PROCESS of DIRECT transfer of heat through matter due to temperature difference. When two objects are in direct contact with one another, heat will be transferred from the hotter object to the COLDER one. The objects which allow heat to travel easily through them are called conductors.
Convection: Convection is the process in which heat transfer is by actual movement of molecules in fluids such as liquids and gases. In convection, molecules move freely from one place to another.
14372.

Two masses 10 gm and 40 gm are moving with kinetic energies in the ratio 9 : 25. The ratio of their linear momenta is,

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`5:6`
`3:10`
`6:5`
`10:3`

ANSWER :B
14373.

A plane isin level flight at constant speed and each of its two wings has an area of 25m^(2) If the speed of the air is 180/h over the lower wingand 234km/h over the upper wing suface , determine the plane's mass . (Take air density to be 1kgm^(-3)).

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<P>

Solution :Speed of air over lower WING `v_(1)=180km//`
`=(180xx1000)/(3600)`
`=50ms^(-1)`
and speed over the upper wing`v_(2)=234km//h`
`=(234xx1000)/(3600)`
`=65ms^(-1)`
Bernoulli.s equation,
`P_(1)+(1)/(2)rhov_(1)^(2)+h_(1)rhog=P_(2)+(1)/(2)rhov_(2)^(2)+h_(2)rhog`
but `h_(1)=h_(2)`
`thereforeP_(1)-P_(2)=(1)/(2)rho(v_(2)^(2)-v_(1)^(2))`
`=(1)/(2)xx1[65^(2)-50^(2)]`
`=(1)/(2)[4225-2500]`
`=(1)/(2)[1725]`
`thereforeP_(1)-P_(2)=862.5Pa`
RESULTANT force on wings `F=(P_(1)-P_(2))A`
`=862.5xx2xx25`
Two wings)
`=43125N`
Now , F=mg
`thereforem=(F)/(G)=(43125)/(9.8)`
`thereforem=4400.5kg`
`thereforem=4393kg`
14374.

Assertion : Work and heat are two equivalent form of energy Reason: Work is the transfer of mechanical energy irrespective of temperature difference, whereas heat is the transfer of thermal energy because of temperature difference only.

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Both ASSERTION and REASON are TRUE and the reason is the CORRECT EXPLANATION of the assertion.
Both assertion and reason are true but reason is not the correct explanation of the assertion
The assertion is true but reason is false.
Assertion and reason both are false.

Answer :A
14375.

A body executing forced oscillations under driving is in sharp resonance. If damping increses then sharpness of resonance.

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decreases
remains same
increases
first INCREASE then decreases

Answer :A
14376.

Consider two observers moving with respect to each other at a speed v along a straighline. They observe a block of mass m movig a distasnce l on a rough surface. The following quantities will be same as observed by the two observers

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kinetic ENERGY of the BLOCK t tiem t
work DOEN by friction
tota work done the block
accelertion of the block

Answer :D
14377.

Physics is more of a philosophy,nay more of a methematical science. Which is true ?

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SOLUTION :PHYSCIS is a beautiful COMBINATION of PHILOSOPHY
and a methematical science.
14378.

Two satellites A and B of the same mass are orbiting the earth at altitudes R and 3R, respectively, where R is the radius of the earth. Let their orbits are circular K and U represent kinetic and potential energies, respectively. Choose the correct option (s )

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`(K_(A))/(K_(B)) =2`
`(U_(A))/(U_(B)) =2`
`(K_(A))/(K_(B)) = (1)/(2)`
`(U_(A))/(U_(B)) = (1)/(2)`

ANSWER :A::B::C
14379.

A spinning cricket ball in air does not follow a parabolic trajectory

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SOLUTION :A SPINNING cricket ball would have followed a parabolic trajectoryhas there been no air .
But because of air the MAGNUS EFFECT TAKES place . Hence , the spinning cricket ball deviates from its parabolic trajectory .
14380.

A vehicle of mass M is accelerated on a horizontal firctionless road under a force chanign its velocity from u to v in distance. A costant power P is given by the engine of the vehicle, then

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`V=[(3PS)/(2M)+u^(2)]^(1//2)`
`V=[(2PS)/(3M)+u^(2)]^(1//2)`
`V=[(3PS)/M+u^(3)]^(1//3)`
`V=[(3PS)/(2M)+u^(3)]^(1//3)`

Answer :C
14381.

Give information for the units of pressure torr and bar .

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Solution :A pressure EQUIVALENT of 1 mm -Hg is CALLED a torr (after Torricelli):1 torr `=133P_(a)`
The mm of Hg and torr are USED in medicine and physiology.
The bar and millibar are used in meteorology. `1"bar"=10^(5)P_(a)`
14382.

Two wires of same length and thickness are joined end to end. Their Young's moduli are 15 xx 10^10 pa and 20 xx 10^10pa . If the combination is stretched by a certain load, the elongations of these wires will be in the ratio

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`3:4`
`4:3`
`1:1`
`1:2`

ANSWER :2
14383.

A particle with position vector has a linear momentum p. Which of the following statements is true in respect of its angular momentum L about the origin?

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L acts along p
L acts along R
L is maximum when p and r are parallel.
L is maximum when p is perpendicular to r.

Solution :Angular MOMENTUM `VECL` is given as `vecL = vecr xx vecp = rp sin theta`
where `, vecr`= POSITION vector of the particle w.r.t. origin `vecp`= its linear momentum .
`vecr xx vecp` is maximum when `vecp` is perpendicular to `vecr ` i.e., `theta = 90`
14384.

The sign of work done by a force on a body is important to understand. State carefully if the following quantities are positive or negative : Work done by a man in lifting a bucket out of a well by means of a rope tied to the bucket.

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SOLUTION :POSITIVE
14385.

The sign of work done by a force on a body is important to understand. State carefully if the following quantities are positive or negative : Work done by gravitational force in the above case

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SOLUTION :NEGATIVE
14386.

The best laboratory approximation to an ideal black body is

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A LUMP of CHARCOAL heated to HIGH temperature
A glass surface coated with COALTAR
A metal coated with a black dye
A hollow enclosure BLACKENED inside with soot and having a small hole

Answer :D
14387.

Fig show two bodies A and B of masses 2.5 kg and 2.8 kg respectively from a rigid support by two inextensible wires each of length 1.8 m . The upper wire is of negligible mass and lower wire is of mass 1.5 kg // m . If the entire system moves upwards with an acceleration of 2 m//s^(2), find tension (i) at middle point p of upper wire (ii) at middle point Q of lower wire . Take g =10 m//s^(2) .

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<P>

Solution :As the SYSTEM moves upwards with an acceleration ` a = 2 m// s^(2) `
`:.` tension in the wire `T= R = m (g+a)`
(i) For point P at point of upper wire
`m= m_(A) + m_(B)` + mass of lower wire
`= 2.5 + 2.8 + 1.8 xx 1.5 = 8.0 kg`
`:. T = 8 (10+2) = 96 N`
(ii) For point Q mid point of lower wire
`m = m^(B)` + mass of half LENGTH of lower wire
`= 2.8 + (1.8)/(2)xx 1.5 = 4.15 kg `
`:. T = m (g+a) = 4.15 (10+2) = 49.8 N` .
14388.

A constant power .P. is applied to a particle of mass .m.. The displacement of the particle when its velocity increases from v_(1) to v_(2) is (ignore friction)

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SOLUTION :Power P = F.v = (ma) v = `a = (P)/(mv) RARR v.(dv)/(ds) = (P)/(mv)` `v^(2).dv=(P)/(m)ds "" (P)/(m)UNDERSET(0)overset(s)intds=underset(v_(1))overset(v_(2))intv^(2).dv`
`(P)/(m)s=(1)/(3)(v_(2)^(3)-v_(1)^(3))"" therefores=(m)/(3p)(v_(2)^(3)-v_(1)^(3))`
14389.

(A): It is possible to transfer heat energy frombody at lower temperature to body at higher temperature by using externalagency (R) : External agency extracts larger heat from colder body by doing work rejects smaller heat to hotter body, hence heattransfer is possible.

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If both (A) and (R) are TRUE and (R) is thecorrect EXPLANATION of (A)
If both (A) and (R) are true and (R) is NOTTHE correct explanation of (A)
If (A) is true but (R) is FALSE
If (A) is false but (R) is true

Answer :C
14390.

A fresh air is composed of ntirogen N_(2)(78%) and oxygen O_(2)(21%). Find the rms speed of N_(2)andO_(2) at 20^(@)C.

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Solution :Absolute temperature `T=20^(@)C+273=293K`
Gas CONSTANT `R=8.32J mol^(-1)K^(-1)`
For, Nitrogen `(N_(2)`,
Molar mass (M) `=28 g PER mol =28xx10^(-3)kg/mol`
`therefore nu_(rms)=sqrt((3RT)/M)=sqrt((3xx8.32xx293)/(28xx10^(-3)))=sqrt((7313.28)/(28xx10^(-3)))`
`(nu_(rms))_(N_2)=511ms^(-1)`
For, Oxygen `(O_(2))`,
Molar mass (M) =32 g per mol `=32xx10^(-3) kg//mol`
`therefore nu_(rms)=sqrt((3RT)/M)=sqrt((3xx8.32xx293)/(32xx10^(-3)))=sqrt((7313.28)/(32xx10^(-3)))`
`(nu_(rms))_(O_2)=478ms^(-1)`
14391.

A glass tube is attached to the bottom of a thin iron rod of length 1m. To what height must the glass tubebe filled with mercury so that the centre of mass of the pendulum formed will not rise or all with changes in temperature? (Linear expansivity of glass =9xx10^(-6)K^(-1), that of iron =12x10^(-6)K^(-1) and cubical edxpansivity of mercury =18.2xx10^(-5)K^(-1))

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ANSWER :`0.146m`
14392.

A body slides down a smooth plane starting from rest in 4sec. Time taken to slide first 1/4 of the distance is

Answer»

2S
1s
3s
(1/8)s

Answer :A
14393.

Find the r.m.s velocity of nitrogen molecules at 15^0 C and 76 cm of Hg.

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SOLUTION :`5.08xx10^2` m/s
14394.

Explain the characteristics of elastic and inelastic collision.

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Solution : In any collision process, the total linear momentum and total energy are always CONSERVE whereas the total KINETIC energy need not be conserved always Some part of the into kinetic energy is transformed to other forms of energy. This is because the impact collisions and deformation occurring due to collisions may in general, produce heat, sound light to By taking these effects into account, we classily the types of collisions as follows
(a) Elastic collision (b) Inelastic collision
(a) Elastie collision: In a collision, the total initial kinetic energy of the bodies before collision is EQUAL to total final kinetic energy of the bodies (after collision) then is called as elastic collision i.e.
Total kinetic energy before collision=Total kinetic energy after collision
(b) Inelastie collision in a collision, the total initial kinetic energy of the bodies (before collision is not equal to the total final kinetic energy of the bodies after collision) it is called as inelastic collision i.e.,
Total kinetic energy before collision ne Total kinetic energy after collision
`("Total kinetic energy before collision")-("Total kinetic energy after collision")=("loss in energy during collision")=TRIANGLEQ`
Even though kinetic energy is not conserved but the total energy is conserved. This is because the total energy contains the kinetic energy term and also a term `triangleQ`, which includes all the losses that take place during collision. Note that loss in kinetic energy during collision is transformed to another form of energy like sound, THERMAL, etc. Further, if the two colliding bodies stick together after collision such collisions are known as completely inelastic collision or perfectly inelastic collision. Such a collision is found very often. For example when a clay putty is thrown on a moving vehicle, the clay putty for (or Bubblegum) stricks to the moving vehicle and they move together with the same velocity.
14395.

If L = 2.06 cm pm 0.02 cm , B = 1.11 cm pm 0.03 cm . What are (L+B) and (L-B) equal to?

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Solution :`L+B= 3.17cm +- 0.05cm`
`L-B= 0.95 cm +- 0.05cm`
Please NOTE that ACTUAL values i.e. 2.06 cm and 1.11 cm are ADDED in CASE of (L+B) and SUBTRACTED in case of (L-B), but absolute errors are added in both cases.
14396.

A thin uniform circular disc of mass M and radius R is rotating in a horizontal plane about an axis passing through its centre and perpendicular to its plane with an angular velocity omega Another disc of same thickness and radius but of mass (1)/(8)M is placed gently on the first disc co-axially. The angular velocity of the system is now

Answer»

`(8)/(9) OMEGA`
`(5)/(9) omega`
`(1)/(3) omega`
`(2)/(9) omega`

ANSWER :A
14397.

Area of surface of a person is "1.9 m"^(2) and its body temp. is 37^(@)C and room temp. is 22^(@)C. Temperature of skin is 28^(@)C, then find the rate of emission of heat. Emissivity of skin is 0.97.

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Solution :Rate of energy emission of body,
`H=Aesigma(T^(4)-T_(S)^(" 4"))""T=273+28=301K`
`""T_(S)=273+22=295K`
`1.9xx0.97xx5.67xx10^(-8)[(301)^(4)-(295)^(4)]`
`=10.44981xx[8208541201-7573350625]xx10^(-8)`
`=10.44981xx635190576xx10^(-8)`
`=66.37620832`
`~~66.4W`
This rate is greater than half of rate of energy produced on body (120 W) under thermal steady state.
To decrease this LOSS, in modern arctic clothes, there is a small smooth excessive METAL layer which reflects the emitter energy from body hence loss DECREASES.
14398.

Two equal uniform rods P and Q each of length l move with the same velocity v as shown in the figure. The second rod has an angular velocity omega(lt6v//l" and clockwise") about its centre of gravity g^(1) in addition to v.

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<P>If the ENDS A and A' are SUDDENLY fixed simultaneously, both the RODS will start to rotate with the same angular velocity
If the ends A and A' are suddenly fixed simultaneously, the rod Q will start to rotate with greater angular velocity
If the ends B and B' are suddenly fixed simultaneously,bothrods will start to rotate with the same angular velocity
If the ends B and B' are suddenly fixed simultaneously, the rod P will start to rotate with greater angular velocity

Answer :B::D
14399.

(A): The only difference between mean velocity and mean speed is that mean velocity is a scalar and mean speed is a vector (R): Magnitude of mean velocity of the gas molecules is same as their mean speed.

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Both (A) and ( R) are true and (R ) is the correct EXPLANATION of (A)
Both (A) and (R ) are true and (R) is not the correct explanation of (A)
(A) is true but (R ) is FALSE
Both (A) and (R ) are false

Answer :D
14400.

An equilateral triangle is made from three mass less rods, each of length l. Two point masses m are attached to two vertices. The third vertex is hinged and triangle can swing freely in a vertical plane as shown.It is released the position shown with one of the rods vertical. Immediately after the system is released, find – (a) tensions in all three rods (specify tension or compression), (b) accelerations of thetwo masses

Answer»


Answer :(a) `(5)/(4) mg, (mg)/(2), (3)/(4) mg`
(B) `(sqrt3)/(4) G`