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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.
| 12501. |
Speed of sound wave in air. |
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Answer» is independent of temperature. Density of dry air is less than that of MOIST air. Hence velocity of sound in dry air would be more than that in moist air (humid air). |
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| 12502. |
Two bodies of mass 60 kg and 30 kg moving in the same direction along straight line velocity 40cms^(-1) and 30cms^(-1)respectively suffer one dimensional elastic collision. Find their velocities after collision |
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Answer» Solution :`"Mass "m_(1)=60KG , "Mass" m_(2)=30KG ,v_(1)=40cms_(-1) ,v_(2)=30 cms^(-1)` `v_(1)((m_(1)-m_(2))/(m_(1)+m_(2)))u_(1)+((2m_(1))/(m_(1)-m_(2)))u_(2)` `v_(2)=((m_(2)-m_(1)))/(m_(1)+m_(2))=u_(2)+ ((2M^(2))/(m_(1)+m_(2)))u_(2)` Substituting the values, we get , `v_(1)=(60-30)/90xx40+(2xx30)/90xx30` `v_(1)=1/9[1200+1800]=(39000)/90=33.3cm s^(-1)` likewise, `v_(2)=(30-60)/90xx30+(2xx60)/90xx40` `v_(2)=1/9[1200+18000]=(3900)/90=43.3cm s^(-1)` |
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| 12503. |
What is (i) the absorbing power and (ii) the reflecting power of a 'perfect black body' ? |
| Answer» SOLUTION :(i) 1 (ONE)(II) ZERO | |
| 12504. |
A gardener is watering plants at the rate 0.1 litre/sec using a pipe of cross - sectional area 1cm^(2). What additional force he has to exert if he desires to increase the rate of watering two times? |
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Answer» Solution :`F=Adv^()=((Av)^(2)d)/(A)`. If the rate of watering of PLANT `(Av)` is doubled, it means that the amount of water poured/sec is doubled which is possible only if VELOCITY is doubled. Hence, force is to be made 4 times. `therefore"Additional force = 3 times initial force"` `=3Adv^(2)=3((Av)^(2))/(A)d` `=(3xx0.1xx0.1xx10^(3))/(10^(-4))=3xx10^(5)N` |
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| 12505. |
The power radiated by a black body A is E_(A) and the maximum energy radiated was at the wavelength lambda_(A).The power radiated by another black body B is E_(B) = NE_(A) and the radiated energy was at the maximum wavelength, (3)/(4)lambda_(A). What is the value of N? |
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Answer» Solution :According to Wien.s displacement law `lambda_("MAX")T="constant for both object A and B"` `lambda_(A)T_(A)=lambda_(B)T_(B)` `"Here"lambda_(B)=(1)/(2)lambda_(A)` `(T_(B))/(T_(A))=(lambda_(A))/(lambda_(B))=(1)/((1)/(2))=2` `T_(B)=2T_(A)` From Stefan - Boltzmann law `(E_(B))/(E_(A))=((T_(B))/(T_(A)))^(4)=(2)^(4)=16=N` Object B has emitted at LOWER wavelength compared to A. So the object B would have emitted more energetic radiation than A. |
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| 12506. |
. A lorry and a car moving with same KE are brought to rest by applying the same retarding force. Then |
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Answer» Lorry will COME to REST in a shorter distance |
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| 12507. |
What effect does couple have on a body? |
| Answer» SOLUTION :A couple have a curning EFFECT, but no resultant FORCE ACTS on a BODY. | |
| 12508. |
A neutron travelling with a velocity v and kinetic energy E collides head-on elastically with a nucleus of mass number A at rest. Calculate the fraction of total energy retained by the neutron. |
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Answer» Solution :`v_1 = ((m_1 - m_2)u_1)/((m_1 + m_2))` while `u_2 = 0` `:. v_1 = ((1 - A)u_1)/((1 + A))` Neutron is `""_(0)n^(1)`. Its MASS number = 1. FRACTION of ENERGY RETAINED by neutron `= ((1/2 mv_1^2)/(1/2 m u_1^2)) = ((v_1)/(u_1))^(2) = ((1 - A)/(1 + A))^(2) = ((A -1)/(A + 1))^(2)` . |
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| 12509. |
(A) : Two circular discs of equal masses and thickness made of different material, will have same moment of inertia about their central axes of rotation.(R ) : Moment of inertia depends upon the distribution of mass in the body. |
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Answer» Both (A) and (R ) are TRUE and (R ) is the CORRECT explanation of (A) |
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| 12510. |
A uniform disc of mass M and radius R is rotating freely about its central vertical axis with angular speed w0. Another disc of mass m and radius r is free to rotate about a horizontal rod AB. Length of the rod AB isL (lt R) and its end A is rigidly attached to the vertical axis of the first disc. The disc of mass m, initially at rest, is placed gently on the disc of mass M as shown in figure. Find the time after which the slipping between the two discs will cease. Assume that normal reaction between the two discs is equal to mg. Coefficient of friction between the two discs is mu. |
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Answer» |
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| 12511. |
A string will break under a load of 5 kg. To the one end of such a 2m long string a mass of 1kg is attached. The maximum rpm in the horizontal plane so that the string does not break is (g=10 ms^(-2)) |
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Answer» `28.66` |
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| 12512. |
The density of a non - uniform rod of length 1 m is given by lambda (x)=p(1+q x^(2)) where p and q are constants and0 le x le 1. The centre of mass of the rod will be at ……… |
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Answer» `(3)/(4)[(2+q)/(3+q)]` |
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| 12513. |
Two slabs A and B of eqal surface area are placed one over the other such that their surfaces are completely in contact. The thicknness of slab A is twice tthat of B. The coefficient of thermal conductivity of slab A is twice that of B. The first suface of slab A is maintained at 100^(0)C, while the second surface of slab B is maintained at 25^(0)C. The temperature at the contact of their surfaces is |
| Answer» Answer :C | |
| 12514. |
SHO of periodic time 2 second starts its oscillation from the lower end of its path of motion, its phase will be……….at time 2 second. |
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Answer» Solution :`implies THETA= (2pi t)/(T)+PHI = (2pi)/(2)xx2+ (3PI)/(2)` `=2pi + (3pi)/(2) = (7pi)/(2) rad`. |
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| 12515. |
Two bodies of masses m_(1) and m_(2) are moving with velocity v_(1) and v_(2) respectively in the same direction. The total momentum of the system in the frame of reference attached to the centre of mass I (v is relative velocity between the masses) |
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Answer» `(m_(1)m_(2)V)/(m_(1)-m_(2))` |
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| 12516. |
A 75kg man stands in a lift . What force does the floor exert on him when the elevator starts moving upward with an acceleration of 2ms^(-2) Given : g=10ms^(-2). |
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Answer» Solution :`R-mg=ma, R=mg+ma=m(g+a)` `=75(10+2)N=900N` `=(900)/(10) KG wt.=90kg .wt.` |
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| 12517. |
A particle executes simple harmonic motion with an angular velocity and maximum acceleration of 3.5 rad // sand 7.5 m // s^(2) respectively. Amplitude of the oscillation is .......... . |
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Answer» 0.36 `:. a=(d^(2)x)/(DT^(2))=-Aomega^(2) sin omegat` `:."Maximum acceleration" |a_(MAX)|=Aomega^(2)` Now `Aomega^(2)=7.5` `A=(7.5)/(omega^(2))=(7.5)/((3.5)^(2))=0.61` |
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| 12518. |
Which of the following graphs correctly represent the variation of beta = -(dV//dP)/(V)with P for an idealgas at constant temperature? |
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Answer»
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| 12519. |
Identify the vector quantity among the following: |
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Answer» distance |
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| 12520. |
The electric energy consumed by a 500 W bulb in 2 hours is …….. unit . |
| Answer» Solution :UNIT `= (Pxxt)/1000 = (500xx2)/1000 = 1 ` . | |
| 12521. |
Which gas has higher specific heat capacity, a monatomic or diatomic gas at room temperature ? |
| Answer» SOLUTION :3R/2 for MONOATOMIC and SR/2 for DIATOMIC. i.e., for diatomic it is GREATER. | |
| 12522. |
A particle moves so th at its p o sitio n vector isomega Whereis a constant which of the following is true |
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Answer» velcocity and acceleration both are perpendicular to `vecr` Here`vec(r )=COS epsilonhat( X)=- epsilont hat( x) +epsilonhat(y )` Acceleration`vec( a) = (d v)/(d t)=- epsilon^(2)cost hat( x )` `vec( a)= - epsilon ^(2) [ cosepsilon t hat( x)+SIN epsilonhat(y )` Here`vec( v) . vec( r ) =cos epsilon t hat( x ) +sinepsilon hat( y )` `=- epsiloncosepsilon TSIN epsilont +epsilont sin epsilon t` `vec( v )_|_ vec( r )` |
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| 12523. |
Assuming the earth's orbit around the sun to becircular, show that the area swept byits radius vector in unit time (areal velocity of the earth) is a constant. |
Answer» Solution :Suppose the earthmoves in a circular orbit of radius r with the sun at the centre (O) of the orbit[Fig. 1.5].  Let the radius vector OP , in an infinitesimal interval of TIME dt, describe an angle `d""THETA` at the centre. Hence are PQ =`rd""theta`. As value of PQ is very small, the arc PQ can be taken to be a straight line (chord PQ). `therefore` Area swept in time dt =area of TRIANGLE OPQ `=1/2 OP xxPQ =1/2r*rd ""theta=1/2r^2d""theta` `therefore` Area swept per unit time `=1/2 r^2*(d""theta)/(dt)=1/2 r^2omega` = constant [as orbit is circular, r and `omega` of the earth should be constants] `therefore` Area swept in unit time by the radius vector of the earth is a constant. |
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| 12524. |
One metre how many light years |
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Answer» `1.057 XX 10^(-15) 1y` |
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| 12525. |
In our solar system, asteriods, small satellites, comets with diameters less than 600 km can be very irregular in shape, where as those with larger diameters are spherical. Only if the rocks have sufficient strength to resists gravity then an object can maintain a non-spherical shape. In solar system, there is a planatary object of uniform density rho and radius R. Find the compressive stress S (defined as force per unit cross-sectional area ) near the centre of the planetary object |
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Answer» `S = piGrho^(2)R^(2)` |
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| 12526. |
In our solar system, asteriods, small satellites, comets with diameters less than 600 km can be very irregular in shape, where as those with larger diameters are spherical. Only if the rocks have sufficient strength to resists gravity then an object can maintain a non-spherical shape. In solar system, there is a planatary object of uniform density rho and radius R What is the largest possible size of a non-spherical self gravitating satellite made of concrete ? Assume that concrete has maximum compressive stress of4. 0 xx 10^(7)N//m^(2) and a density rho = 3000 kg//m^(3) |
| Answer» ANSWER :C | |
| 12527. |
From the given P-T graph of CO_2you may concluded that it will be |
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Answer» Vapour at `-76^@C` under 1 atm |
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| 12528. |
A solid sphere is moving on a horizontal plane. Ratio of its translational K.E. and rotational kinetic energy is [Hint : E_L/E_r = (1/2mv^2)/(1/2Iomega^2) = (mv^2)/(2/5mr^2 omega^2) = (mv^2)/(2/5mv^2)=5/2] |
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Answer» `1/5` |
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| 12529. |
A non-homogeneous sphere of radius R has the following density variation : rho{{:(rho_(0),rleR//3),(rho_(0)//2,(R//3) lt r le (3R//4)),(rho_(0)//8,(3R//4) lt r le R):} The gravitational field at a distance 2R fromthe centre of the sphere is |
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Answer» `0.1piGRrho_(0)` `E=(GM)/((2R)^(2))` where M is the mass of the WHOLE sphere.  Here, `M=(4)/(3)pi((R)/(3))^(3) rho_(0)+{(4)/(3)pi((3)/(4)R)^(3)-(4)/(3)pi((R)/(3))^(3)}(rho_(0))/(2)+{(4)/(3)PIR^(3)-(4)/(3)pi((3)/(4)R)^(3)}(rho_(0))/(8)` `=(4)/(3)piR^(3)rho_(0){(1)/(27)+(27)/(128)-(1)/(54)+(1)/(8)-(27)/(512)}` `=(4)/(3)piR^(3)rho_(0){(512+2916+256+1728-729)/(13824)}` `=(4)/(3)piR^(3)rho_(0){(5156-985)/(13824)}=(4)/(3)piR^(3)rho_(0){(4171)/(13824)}` `=0.402piR^(3)rho_(0)` `:. E=(G(0.402piR^(3)rho_(0)))/((2R)^(2))=0.1piGRrho_(0)` |
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| 12530. |
A metal is heatedfrom 0^(@)C to 500^(@)C and its density reduces to (1)/(1.027) of its original density. Determine the coefficient of linear expansion for this metal, considering it constant for the given range of temperature. |
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| 12531. |
A spark is produced, when two stones are struck against each other. Why? |
| Answer» Solution :The WORK DONE in striking the two STONES against each other gets conerted into heat. This appears as a spark. | |
| 12532. |
The fundamental frequency of a sonometer wire increases by6 Hz if its tension is increased by44 % , keeping the length constant . Find the change in the fundamental frequency of the sonometer wire when the length of the wire is increased by 20 % , keeping the original tension in the wire constant. |
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Answer» Solution :Fundamental frequency of vibrations of string ` n = (1)/( 2 L) SQRT ((T)/(m))` At constant length , `n PROP sqrt(T) or (n)/ ( sqrt(T)) = constant` i. New tension , `T' = T + ( 44)/( 100) T = 1.44 T` If `n'` is new frequency , then `(n')/(sqrt( T')) = (n) /( sqrt(T))` ` n' = ( sqrt ((T')/(T)))n`(ii) ` :.n' = n + 6 ` From Eq. (iii) ` n + 6 = ( sqrt((1.44 T)/(T)))n` or ` n + 6 = 1.2 n` ` 0.2 n = 6 or n = (6)/( 0.2) = 30 Hz` ii.As tension is constant ` n prop (1)/(l) or nl = constant `(iii) When length increase by ` 20%` New length ` l' = l + (20)/( 100) l = 1.2 l` As ` nl= constant ` Therefore , ` nl = n' l'` ` n' = (l)/( l') n = (l)/( 1.2 l) xx 30 = 25 Hz` Change in fundamental frequency ` Delta n =n' - n = 25 - 30 = - 5 Hz` Therefore , ` DELTAN = 5 Hz( decrease)` |
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| 12533. |
A uniform cubical block of mass M and side length L is lying on the edge of a rough table with (1)/(4) th of its edge overhanging. When a small blockof mass m is placed on its top surface at the right edge (see fig.), the cube is on verge of toppling. The block of mass m is given a sharp horizontal impulse so that it acquires a velocity towards B. The small block moves on the top surface and falls on the other side. What is maximum coefficient of friction between the small block and the cube so that the cube does not rotate as the block moves over it. Assume that the friction between the cube and the table is large enough to prevent sliding of the cube on the table. |
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| 12534. |
All the particles of a body are situated at a distance of X from origin. The distance of the center of mass from the origin is |
| Answer» Answer :B | |
| 12535. |
A ring and a disc of same mass roll without slippling along a horizontal surface with same velocity . It the K.E of ring is 8J, then that of disc is |
| Answer» ANSWER :C | |
| 12536. |
Velocity of a particle moving in a stright line varies with its displacement as v=(sqrt(4+4s))m//s. Find its displacement in the first 2 seconds of its motion |
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Answer» Solution :SQUARING the GIVEN equation, we GET `V^(2)=4+4s` Now , comparing it with `v^(2)=u^(2)+2as` we get, `u=2m//s "and" a=2m//s^(2)` THEREFORE , Dispalcement `t=2s`is `s=ut+1/2at^(2)"or"s=(2)(2)+1/2(2)(2)^(2)"or"s=8m` |
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| 12537. |
An open u-tube of uniform cross-section contains mercury. When 27.2 cm of water is poured into one limb of tube (a) how high does the mercury rise in other limb from its initial level ? |
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| 12538. |
Can we use a pendulum watch in an artificial satellite? |
| Answer» Solution :No, in an artifical satellite.a BODY is in a STATE of weoghtleness.i.e.g=0.thereforeT=`2pisqrti/g=infdinate` Inside the satellite the pendulum does not osillate.Hence a pendulum WATCH cannot be used in an artificial satellite. | |
| 12539. |
If velocity of the particle is given by V=sqrt(x) where 'x' denotes the position of theparticle. And initially particle wasat x=4, then which of the following are correct |
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Answer» at `t=2` sec `x=9` |
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| 12540. |
A block is projected with a speed v_0 such that it strikes the point of projection after describing the path as shown by the dotted line. If friction exists for the parth of length d and the vertical circular path is smooth, assuming mu=coefficient of friction, a. Find v_0. b. What is the minimum value of v_0? |
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Answer» `t=sqrt((2[2R])/(G))=sqrt((4R)/(g))` `v_2=d/t=dsqrt((g)/(4R))` From B to C, `1/2mv_1^2=1/2mv_2^2+mg(2R)` `impliesv_1^2=v_2^2+4gR=(d^2g)/(4R)+4gR` From A to B: `v_1^2=v_0^2-2mugdimpliesv_0^2=v_1^2+2mugd` `impliesv_0=sqrt((d^2g)/(4R)+2g[mud+2R])` b. For minimum `v_0`, `v_1=sqrt(5gR)` 
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| 12541. |
A vertical glass capillary tube of length l, radius r closed at its opere end just touches the liquid and then pushed into the liquid so as to equalize levels of liquid both inside & outside of immersion of the tube {ST : T, Paton = P_(0)} |
Answer» SOLUTION :APPLY Boyle.s LAW 
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| 12542. |
(A) : During an adiabatic compression temperature of an ideal gas increase (R ) : Internal energy of a gas can be changed even without addition or withdrawn of heat but by doing work |
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Answer» Both (A) and(R ) are TRUE and (R ) is the CORRECT explanation of (A) |
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| 12543. |
Law of gravitation is not applicable if A)Velocity of moving objects are comparable to velocity of lightB) Gravitational field between objects whose masses are greater than the mass of sun. |
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Answer» A is true, B is false |
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| 12544. |
As a transverse wave strikes against a fixed end ………….. |
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Answer» its PHASE CHANGES by `180^(@)` , but velocity does not change |
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| 12545. |
The liquid drop of density rho, radius r and surface tension o oscillates with time period T. Which of the following expression for T^2 is correct |
| Answer» Answer :A | |
| 12546. |
Column I describles some situtations in which a small object moves. Column II describes some characteristics of these motions. Match the situations in Column I with the characteristics in Column II. |
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Answer» <P> B `to` (p, r) C `to` (S, t) D `to` (q, u) |
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| 12547. |
All physical quantities can be expressed in terms of dimension. Write the physical quantities of the following dimensions [M^1L^1T-1] |
| Answer» SOLUTION :momentuni or IMPULSE | |
| 12548. |
All physical quantities can be expressed in terms of dimension. Write the physical quantities of the following dimensions [M^1L^2T-2] |
| Answer» SOLUTION :WORK of ENERGY or TORQUE | |
| 12549. |
Which one of the fallowing fundamental farces in nature binds protons and neutrons in a nucleus? |
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Answer» GRAVITATIONAL FORCE FARCE |
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| 12550. |
The relative velocity of two consecutive layers is 8cm/s. If the perpendicular distance between the layers is 0.1 cm, then the· velocity gradient will be |
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Answer» 8 `"sec"^(-1)` |
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