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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.
| 14801. |
At which place on the surface of the earth the value of g is maximum ? Gives its reasons. |
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Answer» SOLUTION :`implies` It is MAXIMUM at the poles, because, (i) The radius of EARTH is slightly less at poles. (II) There is no CENTRIFUGAL force act on polar region. |
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| 14802. |
which of the following cylindrical rods will conduct most heat,when their ends are maintained at the same steady temperature? |
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Answer» length 100cm,RADIUS 1 cm |
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| 14803. |
Distance to the moon from the earth is 3.84xx10^8m and the time period of the moon's revolution is 27.3 days. Obtain the mass of the earth. |
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Answer» Solution :`T^2` = `((4pi^2)/(GM))r^3 , M = `(4pi^2)/(T^2G) r^3` = `(4xx(3.14)^2 XX (3.84xx10^8)^3)/((27.3 xx 24 xx 60 xx 60)^2 xx 6.67 xx 10^(-11) M = `6.48xx10^(23)kg` |
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| 14804. |
If a stone dropped from the top of a tower travels half of the height of the tower during last second of its fall, the time of fail is ( in seconds) |
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Answer» `3+sqrt(2)` |
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| 14805. |
Which nuclear process occur in atom bomb ? Or write the principleof 'Atom Bomb' ? |
| Answer» SOLUTION :NUCLEAR FISSION . | |
| 14806. |
Statement I : When there is no relative velocity between source and observer, then observed frequency is the same as emitted Statement II :Velocity of sound when there is no relative velocity between source and observer is zero . |
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Answer» STATEMENT I is TRUE , statement II is true , statement II is a CORRECT EXPLANATION for statement I |
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| 14807. |
Derive the relation between rotational KE and angular momentum. |
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Answer» SOLUTION :Let a rigid body of moment of inertia I rotate with angular velocity `omega`. The angular momentum of a rigid body is `L=I omega` The rotationakl KINETIC energy of the rigid body is `KE=1/2 I omega^(2)` By MULTIPLYING the numerator and denominator of the above equation with I, we GET a relation between L and KE as, `KE=1/2 (I^(2)omega^(2))/I=1/2 ((Iomega)^(2))/I` `KE=(L^(2))/(2I)` |
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| 14808. |
A box is placed on the floor of a truck moving with an acceleration of 7 ms^(-2). If the coefficient of kinetic friction between the box and surface of the truck is 0.5., find the acceleration of the box relative to the truck |
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Answer» `1.7ms^(-2)` |
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| 14809. |
When an object of mass m slides on a friction less surface inclined at an angle thetathen normal force exerted by the surface is |
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Answer» `G cos THETA ` |
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| 14810. |
A block of mass m lieson a wedge of mass M. The wedge in turn lies on smooth horizontal surface. Friction is absent everywhere. The wedge block system is released from rest. All situations given in Table-1 are tol be estimated in the duration the block undergoes a vertical displacement 'h' starting from rest. Match the statement in Table-1 with the results in Table-2ltbgt |
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Answer» |
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| 14811. |
If .R. is the radius of the earth and .w. is angular velocity with which it rotates about its axis, the variation in the value of .g. at 45° latitude of the earth stops its rotation will be |
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Answer» increases by `(RW^(2))/2` |
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| 14812. |
Two blocks of masses 10 kg and 30 kg are placed along a vertical line. If the first block is raised through a height of 7 cm, by what distance should the second mass be moved to raise the centre of mass by 1 cm.? |
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Answer» |
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| 14813. |
The ratio of forces between two bodies kept at a certain distance in air to the force between them when kept in water is ....... |
| Answer» Solution :` implies F_("water")`= 1:1 because gravitational FORCE is INDEPENDENT of MEDIUM between them. | |
| 14814. |
Two moles of an ideal gas X occupying a volume V exerts a pressure P. The same pressure is exerted by one mole of another gas Y occupying a volume 2V. If the molecular weight of Y is 16 times the molecular weight of X, find the ratio of the rms speeds of the molecules of X and Y. |
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Answer» Solution :`V_(RMS)=sqrt((3PV)/(n M)) (m/M=n), v_(1 rms)/v_(2 rms)=sqrt((V_(1))/V_2 n_2/n_1 M_2/M_1)` `v_(1 rms)/v_(2 rms) =sqrt((V)/(2V)) (1/2) (16)/1), (v_(1 rms))/(v_(2 rms))=2` |
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| 14815. |
A copper cube of side of length 1 cm is subjected to a pressure of 100 atmosphere. Find the change inits volume if the bulk modulus of copper is 1.4xx10^(11)Nm^(-2) ("1 atm"=1xx10^(5)Nm^(-2)). |
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Answer» `0.7143xx10^(-10)m^(3)` |
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| 14816. |
In Q.94, total distance travelled by the car is: |
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Answer» `(ALPHA + BETA)/((alpha^(2) + beta^(2))) (t^(2))/(2)` |
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| 14817. |
The unit of permittivity of free space, epsilon_(0) is |
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Answer» COULOMB/newton-metre |
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| 14818. |
Different processes are given in Column -I and its reasons are given in Column -II .Match the appropriately . |
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Answer» |
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| 14819. |
If y=sin omegat-cos omega t then S.T. the function represents a SHM. |
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Answer» Solution :GIVEN, `y=sin OMEGA t-cos omegat` i.e. `y=sinomegat-sin((PI)/(2)-omegat)` |
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| 14820. |
A gardener pushes a lawn roller through a distance 20 m. If he applies a force of 20 kg-wt in a direction inclined at 60^@ to the ground, the work done by him is |
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Answer» 1960 J Here, `F = 20 kg wt = 20 xx 9.8 N = 196 N , s = 20 m , theta = 60^@` `:. W = 196 xx 20 xx cos 60^@ = 1960 J` |
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| 14821. |
The correctness of equation can be checked using the principle of homogeneity in dimensions.Using this principle, check whether the equation f=2pisqrt(l/g) is dimensionally correct, where f - frequency, I - length and g - acceleration due to gravity. |
| Answer» Solution :`[f]=[M^0L^0T^-1],[L]=[M^0LT^0],[G]=[M^0LT^-2]` SINCE `[f]ne[SQRT(l/g)],` the equation is WRONG. | |
| 14822. |
A bob of mass 0.1 kg hung from the ceiling of room by a string 2 m long is oscillating. At its mean position the speed of a bob is 1 ms^(-1). What is the trajectory of the oscillating bob if the string is cut when the bob is (i) At the mean position (ii) At its extreme position. |
| Answer» SOLUTION :(i) PARABOLIC (II) VERTICALLY DOWNWARDS | |
| 14823. |
The moment of inertia of a disc of uniform density about on axis coinciding with its diameter …….. |
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Answer» `(2)/(5)MR^(2)` |
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| 14825. |
The snow accumulatedon the wings of an aeroplane may decrease the lift " why ? |
| Answer» Solution :The SHOW ACCUMULATED changes the structure of the wings, and it NEED not act like the AIRFOIL. This reduces the litt | |
| 14826. |
Two blocks of masses 10 kg and 30 kg are placed along a vertical line. If the first block is raised through a height of 7 cm, by what distance should the second mass be moved to raise the centre of mass by 1 cm. |
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Answer» Solution :`m_(1)=10 kg, m_(2)=30 kg` `Deltax_(cm) =(m_(1)(Deltax_(1)) + m_(2)(Deltax_(2)))/(m_(1) + m_(2)), 1=(10(7) + 30(Deltax_(2)))/(10 + 30) rArr Deltax_(2) =-1 cm` Here "-ve" sign indicates that .`m_(2)`. be MOVED in opposite direction, i.e., `m_(2)`is lowerd by 1 cm. |
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| 14827. |
Same torque is applied on a solid cylinder and solid sphere with same mass and radius. Cylinder rotates about its axis and sphere about its diameter. Then in a given time, the angular velocity acquired is |
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Answer» is more for SPHERE |
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| 14828. |
"In scooters more viscous mobile oil is used in summer than in winter". Why? |
| Answer» SOLUTION :When temperature INCREASES, viscosity decreases. So HIGH viscous oil is used in summer | |
| 14829. |
Two containers, each of volume VO' joined by a small pipe initially contain the same gas at pressure P_(0) and at absolute temperature T_(0). One container is now maintained at the same temperature while the other is heated to 2T_(0). Find (1)commom pressure of the gas, and (2) the number of moles of gas in the container at temperature 2T_(0) |
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Answer» Solution :Let `n_(1) and n_(2)` be the initial NUMBER of moles of gas and `n_(1)^(1) and n_(2)^(1)`be the FINAL number of moles of gas in the two CONTAINERS. Then `n_(1) + n_(2) = n_(1)^(1) + n_(2)^(1)` `(or) (P_(0)V_(0))/(RT_(0)) + (P_(0)V_(0))/(RT_(0)) = (PV_(0))/(RT_(0)) + (PV_(0))/(R2 T_(0))` `(or) (2P_(0)V_(0))/(RT_(0)) + (P_(0)V_(0))/(RT_(0)) [ (3)/(2)] rArr P = (4P_(0))/(3)` From gas equation PV = nRT, we have `(4P_(0))/(3) V_(0) = n_(2)^(1) R (2T_(0))` `therefore n_(2)^(1) = (4P_(0)V_(0))/(3R (2 T_(0))) rArr n_(2)^(1) (4P_(0)V_(0))/(6RT_(0)) = (2P_(0)V_(0))/(3RT_(0)) ` |
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| 14830. |
When a spring is compressed by 3cm, the potential energy stored in it is U. When it is compressed fiurther by 3cm, the increase in potential energy is, |
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Answer» 4U |
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| 14831. |
A particle of mass m is attached to a spring of spring constant 'k' and has a natural frequency omega_(o), . An external force F(t) proportional to cos omegat(omega ne omega_(o))is applied to the oscillator. The maximum displacement of the oscillator will be proportional to |
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Answer» `(m)/(omega_(o)^(2)+ OMEGA^(2))` |
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| 14832. |
Calculate the energy consumed in electrical units when a 75 W fan is used for 8 hours daily for one month (30 days) |
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Answer» <P> Solution :POWER, `p=100W`Time of usage, `t=8hourxx30days=240hours` Electrial energy consned is the product of power and time of usage. `"Electrical energy "=power xx "time of usage"=p xx t` `=75 "watt"xx240 "hou"r=18000" watt HOUR"= 18" kilowatt hour"=18kwh` `1 "electrical unit"=1 kWh` `"Electrial energy"=18 unit` |
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| 14833. |
If vec(b) = 3 vec(i) + 4 vec(j) and vec(a) = hat(i) - vec(j) the vector having the same magnitude as that of vec(b) and parallel to vec(a) is |
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Answer» `(5)/(SQRT2) (HAT(i) - hat(J))` |
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| 14834. |
The cycle in the figure followed by an engine made of an ideal gas in ca cylinder with a piston, theheat exchanged by the engine with the surroundings for adiabatic section AB of cycle is (C_(V)=(3)/(2)R) |
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Answer» `(3)/(2)(P_(B)-P_(A))V_(A)` |
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| 14835. |
The Young's modulus of a perfect rigid body is |
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Answer» 1 |
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| 14836. |
Two planets revolve with same angular velocity about a star. The radius of orbit of outer planet is twice the radius of orbit of the inner planet. If T is time period of the revolution of outer planet, find the time in which inner planet will fall into the star. If it was suddenly stopped. |
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Answer» `sqrt((23gR)/11)` CONSIDER an imaginary comet moving along an ELLIPSE. The extreme POINTS of this ellipse are located on orbit of inner planet and the star, sem-major axis of orbit of such comet will be half of the semi-major axis of theinner planet's orbit. ACCORDING to Kepler's law. if `T'` is the time period of the comet. `(T'^(2))/((r//4)^(3))=(T_(1)^(2))/((r//2)^(3))` `T'^(2)=8/64T_(1)^(2)=(T_(0)^(2))/32( :' T_(1)=T_(0)//2)` `T'=T/(4sqrt(2))` `(T'//2)` represents time in which inner planet will FALL into star. `((T')/2)=T/(8sqrt(2))=(Tsqrt(2))/16` 
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| 14837. |
What is the use of calculating percentage error? |
| Answer» SOLUTION :Then the ERRORS made by DIFFERENT observers can be compared. | |
| 14838. |
Match the parameters given in column I and column II correctly. |
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Answer» 1 - (iii), 2- (v), 3 - (II), 4 - (i) |
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| 14839. |
The parallax of a heavenly body measured from two points diametrically opposite on equator of Earth is 2^(1). Calculate the distance of the heavenly body. [Given radius of the Earth = 6400 km][1" = 4.85 xx 10^(-6) rad] |
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Answer» `8.8xx10^(10) m` diameterof earth, ` d=a xx R_(e)= 2xx 6400 xx 10^(3) m` DISTANCEOF theheavenlybodyfrom thecentreof the earth,` r= (d) /( theta) =( 2xx 6400 xx 10^(3))/( ((pi)/( 60xx80)))` ` r= 4.4 xx 10^(10) m` |
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| 14840. |
The temperature of uniformrodoflength Lhavinga coefficientof linearexpansionalpha_(L) ischangedby Delta T . Calculatethe newmomentof inertia of the uniformrodaboutaxispassingthroughitscenterandperpendicularto anaxisof the rod. |
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Answer» Solution :MOMENT of inertia of a UNIFORM rod of MASS and length l about its perpendicular bisector. Moment of inertia of the rod `I=(1)/(12)ML^(2)`  Increase in length of the rod when temperature is increased by `DeltaT`, is given by `L.=L(1+alpha_(L)DeltaT)` New moment of inertia of the rod `I.=(ML.^(2))/(12)=(M)/(12)L^(2)(1+alpha_(L)DeltaT)^(2)` `I.=I(1+alpha_(L)DeltaT)^(2)` |
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| 14841. |
If the value of the universal gravitational constant G starts decreasing very slowly with time, what will be the effect on the motion of the moon around the earth ? Explain clearly . |
Answer» Solution :Let the earth be at O and the moon, while REVOLVING around the earth , reach the pointA [Fig.1.24].  When the value of G remains unchanged, the moon will CONTINUE its motion ALONG path AB , that is to say its orbit will remain unchanged. On the other hand, if suddenly the valueof G falls to zero , the earth will have no attraction on the moon and the moon will fly off along AC, TANGENTIAL to the orbit . if the value of G decreases slowly with time , the path of the moon will be towardsAD,i.e., along some direction between AB and AC . Asa result , the moon will gradually move away from the earth and will take a spiral path as shown in the |
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| 14842. |
The motion of a satellite will be V = Vel of projection |
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Answer» `V lt SQRT(gr)`It will come back to Earth.s surface |
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| 14843. |
A manometer reads the pressure of a gas in an exclosure as shown in fig.10.25(a) When a pump removes some of the gas, the manometer reads as in fig.10.25(b) the liquid used in the manometer is mercury and the atmospheric pressure is 76cm of mercury. Give the absolute and gauge pressure of the gas in the exclosure for cases (a) and (b) in units of cm of mercury. How would the levels change in case (b) If 13.6 cm of water (immiscible with mercury) are poured into the right limb of the manometer? (ignore the small change in the volume of the gas). |
| Answer» Solution :ABSOLUTE pressure =96CM of Hg. Gauge pressure =20 CM of Hg for (a) absolute pressure =58CM of Hg gauge pressure=-18 cm for Hg | |
| 14844. |
A body projected vertically up with a velocity of 20 ms^(-1) returns to the ground with a velocity of 18 ms^(-1) . The maximum height attained by the body is (g=10 ms^(-2)) |
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Answer» 12.1 m |
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| 14845. |
What is meant by conduction ? |
| Answer» Solution :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. | |
| 14846. |
In a vernier callipers n divisions of its main scale match with (n+5) divisions on its vernier scale. Each division of the main scale is of x unit. Using the vernier principle, find the least count. |
| Answer» Answer :C | |
| 14847. |
One litre of oxygen at a pressure of 1 atm and two litres of nitrogen at a pressure of 0.5 atm are introduced into a vessel of volume 1 litre. If there is no change in temperature, the final pressure of the gas in atm is |
| Answer» ANSWER :C | |
| 14848. |
A disc rolls on ground without slipping . Velocity of centre of mass is v. There is a point P on circumference of disc at angle theta . Suppose v_(p) is the speed of this point. Then, match the following the following table. |
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Answer» <P> |
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| 14849. |
The speed of projective at its maximum height is (sqrt(3))/(2) timeits initial speed. If the range of the projective is p times the maximum height attained by it, then p= |
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Answer» `4//3` |
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| 14850. |
A manometer reads the pressure of a gas in an enclosure as shown in the figure. The absolute and gauge pressure of the gas in cm of mercury is (take atmospheric pressure =76 cm of mercury) |
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Answer» 76,20 Pressure head, h=20 cm of mercury Absolute pressure `=P+h=76+20=96 cm` of mercury GAUGE pressure `=h=20 cm` of mercury. |
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