Saved Bookmarks
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.
| 13601. |
According to Hooke's law of elasticity the ratio of stress to strain |
|
Answer» Does not REMAIN constant |
|
| 13602. |
A body of density p floats with a volume V_(s) of its total volume V immersed in one liquid of density rho_(1) and with the remainder of volume V_(2) immersed in another liquid of density rho where.rho_(1) gt rho_(2) find V_(1) and V_(2) |
|
Answer» `V_(1)=V((RHO-rho_(2))/(rho_(1)-rho_(2)))V_(2)=v((rho_(1)-rho)/(rho_(1)-rho_(2)))` |
|
| 13603. |
A stick of square cross-section (5 cm xx 5 cm) and length '4m' weighs 2.5 kg is in equilibrium as shown in the figure below. Datermine its angle of inclination (in degree) in equilibrium when the water surface is 1 m above the hinge. |
|
Answer» |
|
| 13604. |
The angle of incidence for a ray of light at a refracting surface of a prism is 45^@. The angle of prism is 60^@. If the ray suffers minimum deviation through the prism, the angle of minimum deviation and refractive index of the material of the prism respectively are: |
|
Answer» `45^@,SQRT2` |
|
| 13605. |
Calculate the work done in increasing the angular velocity of a wheel of moment of inertia 200 kg m_2 from 20 rev/s to 40 rev/s. |
|
Answer» Solution :Initial ANGULAR velocity `=omega_(0) = 2pi n_(0) = 2pi XX 20` rad/s FINAL angular velocity `=omega = 2pi n = 2pi xx 40` rad/s Moment of intertia `=I =200 kg m^(2)` Work done `W = 1//2 l omega^(2) -1//2 I omega_(0)^(2)` `=1/2 xx 200 [(80 pi)^(2)-(40 pi)^(2)] = 4.73 xx 10^(4)` J |
|
| 13606. |
Two particles executive SHM of the same amplitude and frequency on parallel lines side by side. They cross one another when moving in opposite directions each time their displacement is (sqrt3)/(2) times their amplitude. What is the phase difference between them? |
| Answer» ANSWER :D | |
| 13607. |
A train takes t see to perform journey, if travel for t/n sec with unifornl acceleration then for ((n-3)/(n))t sec with uniform speed v and finally it comes to rest with uniform retardation. Then average speed of train is |
|
Answer» `(3N-2) V/(2N)` |
|
| 13608. |
The volume of the air bubble increases by 10% when it rises from bottom to the surface of a water pond. If the temperature of the water is constant, find the depth of the pond. |
|
Answer» Solution :Volume of the AIR bubble at the bottom of the pond `(V_(1))=V` Volume of the air bubble at the surface of the pond `V_(2)=V times 110/100` Pressure at the bottom of the pond `(P_(1))=(h+H)` of water column where .h. is the DEPTH of the pond and .H. is the atmospheric pressure (on water BAROMETER) Pressure on the surface of the pond `(P_(2))=H` cm of water column From Boyle.s law, `P_(1)V_(1)=P_(2)V_(2)` (or) `(h+H)V=H(V times 110/100) rArr h+H=H times110/100 rArr h=H times 110/100-H=H times 10/100=H/10` `therefore` Depth of the bond = H/10. |
|
| 13609. |
If gammabe the ratio of specific heats of a perfect heals of a perfect gas, the number of degrees of freedom of a molecule the gas is : |
|
Answer» `25/2(GAMMA -1)` |
|
| 13610. |
The correct order in which the dimensions of 'Length" increases in the following physical quantities is A) Permittivity B) Resistance C) Magnetic permeability D) Stress |
|
Answer» A,B,C,D |
|
| 13611. |
An iron block is suspended from a string and is then completely immersed in a container of water. The mass of iron is 1 kg and its density is 7200 kg //m^(3). What is the tension in the string before and after the iron block is immersed ? |
|
Answer» `= 9.8 - (1xx 1000xx 9.8 //7200) = 8.44 N` |
|
| 13612. |
A thin uniform rod of mass 5 kg and length 1 m is held in horizontal position with the help of strings attachedd to ends of rod. Other ends of strings are held by some external agent. Now end A is pulled down with speed v_(A)=3t and end B is pulled down with speed v_(B)=t , where t is time in second. Choose the correct choice(s) |
|
Answer» Angular ACCELERATION of ROD is 2 `"rad"//s^(2)` |
|
| 13613. |
A billiardballof massM ,movingwithvelocityv_1collideswithanotherball of thesamemassbutat rest. If thecollisioniselastic , theangleof divergenceafterthe collisionis |
|
Answer» `0^(@)` |
|
| 13614. |
The percentage change in the time period of a seconds pendulum when its amplitude is reduced by 30% is (assume is very small) |
| Answer» ANSWER :B | |
| 13615. |
A particle is moving along a straight line parallel to x-axis with constant velocity. Its angular momentum about the origin |
|
Answer» decreases with time `x = vt , y = b ` and z = 0 while COMPONENTS of velocity will be `v_(x)= v , v_(y) = 0 ` and `v_(z) = 0` (as it is moving parallel to x-axis)  So , `vecL` So , `vecL = vecr xx vecp = m |{:(hati , hatj , hatk) , (vt , b , 0), (v , 0 ,0):}| = hatk m[vt xx0 - vb] =-mvb hatk` i.e., angular momentum has magnitude mvb and is directed along NEGATIVE z-axis i.e., angular momentum remains constant . |
|
| 13616. |
Moment of inertia of a body comes into play |
|
Answer» in linear MOTION |
|
| 13617. |
Choose the false statement. 1st Law of thermo dynamics |
|
Answer» does not tell us about the efficiency WITHWHICH heat can be converted in to work. |
|
| 13618. |
A lorry and a car moving with same KE are brought to rest by applying the same retarding force. Then |
|
Answer» LORRY will COME to REST in a SHORTER distance |
|
| 13619. |
Which are warmer,two blankets each of half inch thickness or single one of one inch thickness and why? |
| Answer» Solution :TWO blankets are warmer.The air enclosed between two blankets GIVES an additional insulation and prevents the LOSS of heat from our BODY. | |
| 13620. |
No physicist has ever '' seen'' an electron. Yet, all physicits believe in the existence of electron. An intelligent but superstitous man advances this analogyto argue that 'ghosts' exist even though no one has 'seen' one. How will you refute his argument? |
|
Answer» Solution :No physicist has EVERY 'seen' an electron. This is true. But there is so much of EVIDENCE that establishes the EXISTANCE of electrons. On the CONTRARY, there is headly any evidence, direct or indirect to ESTABLISH the existence of ghosts. |
|
| 13621. |
A ball is projected from the ground at a speed of 10 ms^(-1)making an angle of 30^@with the horizontal. Another ball is simultaneously released from a point on the vertical line along the maximum height of the projectile. The initial height of the second ball is (Take g = 10 m s^(-2)) |
|
Answer» 6.25 m `:. H = (10)^(2)xxsin^(2) 30^(@) /(2xx10)=5/4=1.25 m` Time for attaining maximum height, `t = (u sin theta)/g` `:. t = (10 xx sin 30^@)/10 = 0.5 s` `:.`Distance of VERTICAL fall in 0.5 s ,`S = 1/2 "gt"^2` or `S=1/2 xx 10 (0.5)^(2) = 12.5 m` `:.` Initial height of the second ball = S + H = 12.5 + 12.5 = 2.50 m |
|
| 13622. |
When some load is attached to the spring of spring constant then it elongates through 0.8 m. If it is further stretches by 0.05m then the period of oscillation is |
|
Answer» `(2PI)/(7)` sec |
|
| 13623. |
Some of notable matters in derivation of MvecA=vec(F)_(ext). |
|
Answer» Solution :(1) To DETERMINE the motion of the centre of mass no knowledge of internal forces of the system of particles is required for this purpose to know only the EXTERNAL forces. (2) To obtain this equation we did not need to specify the nature of the system of particles. The system may be a collection of particles in which there may be all kinds of internal motion, or it may be a rigid BODY which has either pure translational motion or a combination of translational and rotational motion. (3) WHATEVER is the system and the motion of its individual particles, the centre of mass MOVES according to `MvecA=vec(F)_(ext)`. |
|
| 13624. |
A mass of 1 kg is suspended by a thread. It is lifted up with an acceleration 4.9m/s^(2) (ii) lowered with an acceleration of 4.9 mcdot s^(-2). The ratio of the tension on the thread is |
|
Answer» Solution :Tension while lowering the MASS, `T_(2) = G - a = g - (g)/(2) = (g)/(2)` `therefore""T_(1) : T_(2) = (3g)/(2) : (g)/(2) = 3 : 1 ` Theoption A is correct. |
|
| 13625. |
In poiseuille.s method of determination of coefficient of viscosity, the physical quantity that requires greater accuracy in measurement is h |
|
Answer» pressure difference |
|
| 13626. |
Two identical blocks A and B each of mass M are connected to each other through a ligh string. The system is placed on a smooth horizontal floor. When a constant force F is applied horizontally on the block A, find the tension in the string. |
|
Answer» SOLUTION :The acceleration of the system of two blocks A and `B=("Force")/("Total mass")` `therefore a=(F)/(M+M)=(F)/(2M)` If we CONSIDER the free body diagram of A, the forces ACTING on it are (i) the applied force F and (ii) the tension T on the string as shown in the following fig.  The RESULTANT force `= F-T , Ma = F-T` `M((F)/(2M))=F-T "" (therefore a=(F)/(2M))` `(F)/(2)=F-T`  `T=(F)/(2)` (or) From FBD for B `T=Ma=M(F)/(2M)=(F)/(2)`  `T=(F)/(2)` |
|
| 13627. |
Give Newton's for velocity of sound in air or gases along with the meanings of the symbols used. |
| Answer» SOLUTION :`v = sqrt((p)/(RHO))` where 'p' is PRESSURE on gaseous medium, `rho-` density of gaseous medium. | |
| 13628. |
The minimum velocity required for a body at the bottom of a vertical circle to complete the circle of radius .R. is sqrt(5gR)A body is released from the top edge of an inclined plane as shown in the figure. The minimum value of .h. for which the body can complete the vertical circle is: |
| Answer» ANSWER :D | |
| 13629. |
Young's modulus of rigid body is ……. |
|
Answer» Solution :Infinite. There is no strain in RIGID body. HENCE Young.s MODULUS `= ("stress")/("strain") = ("stress")/(0)`= infinite |
|
| 13630. |
(A): A deforming force in one direction can produce strain in other directions. (R) : The proportionality between stress and strain defines Hooke's law of elasticity. |
|
Answer» Both (A) and (R) are true and (R) is the CORRECT EXPLANATION of (A) |
|
| 13631. |
A simple pendulum of length l carries a bob of mass m. If the breaking strength of the string is mg/2, the maximum angular amplutide from the vertical can be |
| Answer» Answer :C | |
| 13632. |
Oneend of a rodof lengthL andcross - sectioal area A is keptin a furanceof temperature T_(1) . Theotherend of the rod is keptat a tempearute T_(2). Thethermal conducitivity of the material of the rod isK and emissivityof rodis K and emissivityof rodis e. It isgiven that T_(2) = T_(s)+ DeltaT, where Delta T lt lt T_(s), T_(s)beingthe temperature of surroudings. If Delta T prop (T_(1) - T_(s)) , find the proportionalityconstant. Considerthat heat is lostonly by radiation at theendwherethe temperature of the rod is T_(2).. |
Answer» SOLUTION :Rateof Heatgained by rodthroughconduction , `Q = (KA(T_(1) - T_(2)))/(L)`  Rateof Heatlostby rodthrough radiation, `Q_(2) = e A SIGMA (T_(2)^(4) - T_(s)^(4))` In steady state `Q_(1)= Q_(2)` `therefore (KA(T_(1)-T_(2)))/(L) = e A sigma(T_(2)^(4) - T_(s)^(4))..........(1)` Given `T_(2) = T_(s) + DeltaT "" therefore T_(2)^(4) = (T_(s) + DeltaT)^(4) = T_(S)^(4) + (1 + (DeltaT)/(T_(s)))^(4)` As `Delta T lt lt T_(s)` , so usingbionmial theorem . ` T_(2)^(4) = T_(s)^(4) ( 1+(4Delta T)/(T_(s))) rArr T_(2)^(4) - T_(s)^(4) = 4T_(s)^(3) DeltaT ` substitutingthis valuein (1) , we get`(KA(T_(1) - T_(2)))/(L) = e A sigma. (4T_(s)^(3) DeltaT) ` Also`T_(2) = T_(s) + Delta T` `therefore(KA(T_(1) - T_(s) - DeltaT))/(L) = e A sigma (4T_(s)^(3) DeltaT) rArr (K(T_(1) - T_(s)))/(L)- (K)/(L) DeltaT = 4esigma T_(s)^(3) Delta T` (or) `(K(T_(1) - T_(s)))/(L)= (4esigma T_(s)^(3) + (K)/(L)) DeltaT , Delta T = (K(T_(1) - T_(S)))/((4 e sigmaL T_(s)^(3) + K))` Butgiven `Delta T = C(T_(1) - T_(s))` `therefore ` Constant of proportioanlityC is `(K)/(K + 4 e sigma L T_(s)^(3))` |
|
| 13633. |
Which of the following are positive and which are negative work done ? Work done by an applied force on a body moving through a rough horizontal plane with uniform velocity. |
| Answer» SOLUTION :POSITIVE | |
| 13634. |
A cycle followed by an engine (made of one mole of an ideal gas in a cylinder with a piston) is shown in Fig. Find heat exchanged by the engine, with the surrounding for each section of the cycle. AB : Constant volumeBC: constant pressureCD : adiabaticDA: constant pressure. |
|
Answer» <P> Solution :`Q_(AB)=U_(AB)=3/2R(T_(B)-T_(A))=3/2 V_(A)(P_(B)_P_(A))``Q_(BC)=U_(BC)+W_(BC)=(3//2)P_(B)(V_(C)-V_(B))+P_(B)(V_(C)-V_(B))=(5//2)P_(B)(V_(C)-V_(A))` `Q_(CD)=0 Q_(DA)=(5//2)P_(A)(V_(A)-V_(D))` |
|
| 13635. |
Graph of velocity (v) versus time 't' for motion of a motorbike which starts from rest and moves along a stright road is given . (a) Find the average acceleration for the time interval t_(0)=0 to t_(1)=6.0s(b)Estimate the time at which the acceleration has its greatest positive value and the value of the acceleration at the instant. (c) When is the acceleration is zero? (d) Estimate the maximum negative value of the acceleration and the time at which it occurs. |
|
Answer» |
|
| 13636. |
The fractional change in wavelength of light coming from a star is 0.014%. What is the velocity? |
|
Answer» `4.2 XX 10^(3) m//s` |
|
| 13638. |
State difference of mass and weight. |
Answer» SOLUTION :
|
|
| 13639. |
Whatisa deforming force? |
| Answer» Solution :DEFORMING force isa force that produces a change in the size or SHAPE of a BODY. | |
| 13640. |
A wall is of two layers P and Q each made of different materials. Both the layers have same thickness. The thermal conductivity of the material P is twice that of Q. Under thermal equilibrium, the temperature difference across the walls is 36^(@)C. What is the temperature difference across the layer P ? |
|
Answer» SOLUTION :For layer P, `A=A_(1),l=l_(1),K=K_(1)` for layer Q, `A=A_(2),l=l_(2),K=K_(2)` Given `l_(1)=l_(2),A_(1)=A_(2),K_(1)=2K_(2)` We know `Q/t=KA(Deltatheta)/lrArrDeltathetaalpha1/K" "(because"here "Q/t,A&l" are same")` i.e., `(Deltatheta_(P))/(Deltatheta_(Q))=K_(Q)/K_(P)=1/2" "thereforeDeltatheta_(P)=1/3xxDeltatheta=1/3xx36=12^(@)C` |
|
| 13641. |
(A) : A helicopter must necessarily have two propellers. (R ): Two propellers are provided in helicopter in order to conserve linear momentum |
|
Answer» Both 'A' and 'R' and TRUE and 'R' is the correct explantation of 'A' |
|
| 13642. |
A solid block of mass m = 1 kg is resting on a horizontal platform as shown in figure. The z direction is vertically up. Coefficient of friction between the block and the platform is mu = 0.2. The platform is moved with a time dependent velocity givenby vecV = (2thati + thatj+ 3thatk)m//s . Calculate the magnitude of the force exerted by the block on the platform.Take g = 10 m//s^(2) |
|
Answer» |
|
| 13643. |
A block of mass 15 kg is placed on a long trolley. The coefficien t of static friction between the block and the trolley is 0.18. The trolley accelerates from rest with 0.5 ms^(2) for 20 s and then moves with uniform velocity. Discuss the motion of the block as viewed by (a) a stationary observer on the ground, (b) an observer moving with the trolley. |
|
Answer» Solution : Massof blockm =15 kg accelerationof trolley a=0.5 `ms^(2)` (a)WhenblockmoveswithtrolleyForceactingon blockF= m `=7.5 N` Dueto REACTIN forcewill acton blockdirectionof thisforcewill motiontrolley . Staticfriction willact on blockstaticfrictionactingonblock `f= muR` `=mu mg` Heremaximumstaticfriction(26.46 N)`gt ` forceactingon block(7.5 N) ( b )Whenoberseveris movingwithtrolley.thiswill benon-inertiareferenceframe . Henceto observerblockappearto bestationary. |
|
| 13644. |
The position of a particle is given by x = 4 sin 2t, where x is in meters and is in seconds. Find an expression for the acceleration of the particle as a function of time. What is the acceleration at time t=0 ? What is the maximum value of the acceleration? |
| Answer» Solution :`a= (-16 m//s^2 ) SIN 2R , 0 |a_("max")| = 16 m//s^2` | |
| 13645. |
The position of a particle is given by x = 4 sin 2t, where x is in meters and is in seconds. Find an expression for the velocity of the particle as a function of time. What is the velocity at time t = 0 ? |
| Answer» SOLUTION :`V= (8 m//s) COS 2T, 8m//s` | |
| 13646. |
A massive ball moving with speed v collides with a tiny ball by negligible mass. The collision is elastic. Find the speed with which the second tiny ball will move. |
|
Answer» Solution :`V_2= (2(m_1u_1))/((m_1 + m_2))` Since `m_1 > > m_2, (m_1 + m_2) = m_1` or `v_2 = (2m_1v)/(m_1) = 2v` `:.` SNEED of second tiny BALL = 2v. |
|
| 13647. |
Can the whole of work be converted into heat ? |
| Answer» Solution :Yes through THRE fiction work is COMPLETELY convertedto HEAT . | |
| 13648. |
Two perfectly elastic spheres of masses 4kg and 6kgmoving in opposite with velocities of 4 m/s and 6 m/s collide with each other. Their velocites after head on collision are |
|
Answer» 0.4 m/s and -3.6 m/s |
|
| 13649. |
A vector is represented by 3i+2j+2k. It is length in the XOY plane is |
|
Answer» |
|
| 13650. |
A body of mass m is moving along a circular path of radius R such that at any instant the kinetic energy K=K_(0)(t/(t_(0)))^(2) where t_(0) and k_(0) are appropriate constants. Then |
|
Answer» The magnitude of TANGENTIAL component of force acting on it must be constat |
|