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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.
| 14101. |
A stone is dropped into a well and the sound of splash is heard after 5.3sec. If the water is at a depth of 122.5 m from the ground, the velocity of sound in air is . |
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Answer» Solution :If `t_(1)` is the TIME taken by stone to REACH the ground and `t_(2)` the time taken by sound to go up, then `t_(1)+t_(2)=5.33` Since `s=ut+1/2 at^(2)""122.5=0t+1/2 xx 9.8 xx t_(1)^(2)` `t_(1)^(2)=(245)/(9.8)=(2450)/(98)=25""therefore t_(1)=5s` `t_(2)=0.33s` Velocity of sound `=("displacement")/("time")=(122.5)/(0.33)=367m//s` |
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| 14102. |
Some of the dimensional veriables are …………. , ……………. , …………… . |
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| 14103. |
“A seasoned (experienced) cricketer catches a cricket ball coming in with great speed where as a novice (unexperienced) can hurt his hand in same act” - Explain |
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Answer» Solution :` (##KPK_AIO_PHY_XI_P1_C05_E01_068_S01.png" width="80%"> Whenball iscomingto seasonedcricketerhedraws hishandinbackwarddirectionduringcatchingball asshownin figure. THEREBY thetakeslongertime tostopball. thenoviceon theotherhandkeepshishandsprovideand TRIESTO catchinstantly. He need toprodivemuchgreaterimpulseof forceto stopballinstantlyand cangethurt. |
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| 14104. |
A ball is projected vertically down with a velocity of 10 sqrt3 m//s from a height of 20m. On htting the ground if it loses 40% of its energy, the height to which it bounces is (g= 10 ms^(-2)) |
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Answer» 20m |
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| 14105. |
If a satellite is revolving around a planet of mass M in an elliptic orbit of semi-major axis a, show that the orbit al speed of the satellite when it is at a distance r from the focus will be given by v^2=GM(2/r - 1/a). |
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Answer» SOLUTION :Total MECHANICAL ENERGY of the system is `E=-(GMm)/(2a)` which remains conserved `rArr` KE+PE=`-(GMm)/(2a)` At a position .r. obital speed of the SATELLITE is v. The KE=`1/2mv^2` , PE=-`(GMm)/r` So, `1/2mv^2-(GMm)/r=-(GMm)/(2a)` (or) `v^2=GM(2/r-1/a)` . |
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| 14106. |
Is a body in circular motion in equilibrium? |
| Answer» Solution :No a body in circular motion has a CENTRIPETAL ACCELERATION `veca` directed towards the centre of the circle. Since acceleration is not nullified, the body is not in EQUILIBRIUM. | |
| 14107. |
Two particles execute SHM of the same amplitude and frequency on parallel lines side by side. They cross each another when moving in opposite directions each time their displacement is half their amplitude. What is the difference between them ? |
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Answer» `pi/3` SEC |
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| 14108. |
Two plane, smooth surfaces are parallel to each other and are initiallt a distance of 2 metre apart. The two surfaces approach each other with a velocity of 1 cm//sec. A particle starts with a velocity of 4 cm/sec from one surface and collides normally and elastically on the other surface from the time the two surfaces start moving. The collisions continues back and forth till the surfaces touch each other. The total distance covered by the particle is : |
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Answer» 2 m |
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| 14109. |
Column I gives three physical quantities. Select the appropriate units for the choices given in column II. Some of the physical quantities may have more than one choice correct. {:("Column I","Column II"),("1. Inductance","(i) Ohm (second)"),("2. Magnetic induction","(ii) Coulomb"^(2)"(joule)"^(-1)),("3. Capacitance","(iii) Coulomb "("volt")^(-1)),(,"(iv) Newton (ampere metre)"^(-1)),(,"(v) Volt second (ampere)"^(-1)):} |
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Answer» 1 - (iv), 2 - (iii), 3 - (II) and (i) `="volt sec ampere"^(-1)` `="ohm sec."` (ii) `"Magnetic INDUCTION"` `B=(F)/(Il)=("newton")/("ampere metre")` `="newton (ampere metre)"^(-1)` (iii) `"CAPACITANCE"` `C=(Q)/(V)="coulomb volt"^(-1)` `C=(q)/(w//g)=(q^(2))/(w)="coulomb"^(2)" JOULE"^(-1)` |
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| 14111. |
A juggler is showing a trick with n balls. When one of the balls is in his hand all of therest are in flight. If each ball rises to a height of h, then how long does the juggler hold a ball before throwing it up? |
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| 14112. |
In a collinear collision a particle with an intial speed v_(0) strike a stationary particle of the same mass. If the final total kinetic energy is 50 % greater than theoriginal kinetic energythe magnitude of the relative velocitybetween the twoparticles after collision is . |
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Answer» `(v_(0))/4` FINAL Kinetic energy = 50 % of `1/2 mv_(0)^(2) +1/2 mv_(0)^(2)` ` = 1/2 mv_(0)^(2)+1/2 mv_(0)^(2) xx1/2 ` `= 3/4mv_(0)^(2)` . ` :. v_(1)^(2) +v_(2)^(2)=v_(0)` ` :. v_(1)^(2) +2v_(1)v_(2) +v_(2)^(2) =v_(0)^(2) ""....(2)` ` rArr` From ewu . (1) and (2) `2v_(1)v_(2) =v_(0)^(2) -3/2v_(0)^(2) [ :. v_(1)^(2) +v_(2)^(2) =3/2v_(0)^(2)]` ` :. 2v_(1)v_(2) = - (v_(0)^(2))/2 ` and `(v_(1)-v_(2))^(2) = v_(1)^(2)+v_(2)^(2) -2v_(1)V_(2)` `= 3/2 v_(0)^(2) - (-(v_(0)^(2))/2)` ` =3/2 v_(0)^(2) +(v_(0)^(2))/2` ` = 2v_(0)^(2)` ` :. v_(1) - v_(2) = sqrt(2) v_(0)` Relative velocity . |
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| 14113. |
Which property of a metal is manifested when a compressional stress more than the yield point is developed? |
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| 14114. |
A steel wire of length 2 m and 1.2 xx 10^(-7)m^(2) in cross sectional area is stretched by a force of 36 N. Calculate |
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Answer» Solution :stress, strain, increase in length, work done in stretching the wire `(Y=1.8xx10^(11)N//m^(2))` Stress `=("force")/("area")=(36)/(1.2xx10^(-7))=3xx10^(8)N//m^(2)` Stress Y= Strain `:. "Strain" =("stress")/(Y)=(3)/(1.8)xx10^(-3)=1.67xx10^(-3)` INCREASES of length `=Deltal=(l)/("strain")=1.67xx10^(-3)xx2=3.34xx10^(-3)m` Workdone `=(1)/(2)` (stress) (strain) volume `=(1)/(2)xx3xx10^(8)xx(5)/(3)xx10^(-3)(2xx1.2xx10^(-7))=6xx10^(-2)` joule. |
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| 14115. |
what is law of conservation ofangular momentum . |
| Answer» SOLUTION :it states that ,"if no EXTERNAL TORQUE acts ina a SYSTEM its total angular MOMENTUM will remains constatnt ." | |
| 14116. |
During propagation of a plane progressive mechanical wave .......... |
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Answer» all the paricles are vibrating in the same phase.  Above figure shows that at a GIVEN instant, phases of different particles are different. (Because they all have different displacements). Hence, option (A) is false. In case of progressive wave LIKE above, each particle undergoes same magnitude of maximum displacement. Thus, they all have same amplitudes. Hence option (B) is true. Given wave shape is sinusoidal which indicates motion of disturbance of simple HARMONIC type. Hence, all particles perform S.H.M. about their respective equilibrium positions. Thus, option (C) is true. Velocity of wave in a given medium depends on two properties (i) inertia (II) elasticity which ultimately depend on the type of medium. Hence, option (D) is true. |
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| 14117. |
The particle of mass m is projected at t=0 froma point O on the groundwith speed v_(0) at an angle 45^(@)to thehorizontalasshown in Figure 3.44. Computethe magnitude and directionofthe angularmometum of theparticle aboutthe point O atposition : Whenthe velocityof thepaticle is ,vecV= (0,7v_(0))hati-(0.3v_(0))hatj |
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| 14118. |
The pulley A and C are fixed while the pulley B is movable. A mass M_2 is attached to pulley B , while the string has masses M_1 and M_3 at the two ends. Find the acceleration of each mass . |
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Answer» SOLUTION :The force acting on masses ` M_1 , M_2 and M_3` are shown in figure . Let ` a_2 and a_3` be upward accelerationof masses ` M_1 and M_3 and a_2` the downward acceleration of mass `M_2` . For motion of mass `M_1 ` ` T-M_1 g = M_1 a_1 `....... (1) For motion of mass `M_3` `T-M_3g = M_3 a_3` ........ (2) For motion of mass `M_2` `M_2g - 2T= M_2 a_2` ......... (3) Let ` l_1 , l_2 , l_3` be the lengths of vertical portions of the STRING between pulley at any instant. `M_1` goes up through ` x , M_2` goes down through y and `M_3` goes up through z. Then ` l_1 + 2l_2 + l_3 = ( l_1 - x) + 2 (l_2 + y) + l_3 -z` `implies x + z = 2y implies (d^(2) x)/( dt^(2)) + (d^(2) z)/(dt^(2)) = 2(d^(2) y)/(dt^(2))` ` a_1 + a_3 = 2a_2 implies a_2 =(a_1 + a_3)/( 2) `.......... (4) From (1) , ` a_1 =(T)/(M_1) - g ` ....... (5)  From (2) ` , a_3 = (T)/(M_3)-g` ............ (6) substituting ` a_1 and a_3 ` is (4) , we get ` a_2 =(T)/(2) ((1)/(M_1) + (1)/(M_3))-g`......... (7) From (3) , `T=(M_2(g-a_2))/(2)` .......... (8) substituting this value in (7) , we get `a_2 =(M_2)/(4) (g-a_2)((1)/(M_1 ) + (1)/(M_3))-g or a_2 = (M_2 (g-a_2) (M_3+M_1))/(4M_1M_3)-g` ` or 4M_1M_3 a_2 _ M_2(M_1 + M_3) g -M_2(M_1 + M_3) a_2- 4M_1 M_3 g ` `or [4M_1M_3 + M_2(M_1 + M_3)]a_2 = (M_1 + M_3) M_2 g - 4M_1 M_3 g ` ` a_2 = ((M_1 + M_3) M_2 g - 4M_1 M_3g)/(4M_1 M_3 + M_2(M_1 + M_3))` `a_2=(M_1M_2 + M_2 M_3 - 4M_1 M_3)/(M_1 M_2 + M_2M_3 + 4M_1 M_3 )g`...... (9) substituting value of T from (8) in (5) ` a_1 =(M_2(g-a_2))/(2M_1)-g` Substituting value of`a_2` in above equation , we get ` a_1 =(-M_1 M-2 + 3M_2 M_3- 4M_1 M_3)/(M_1 M_2 + M_2M_3 + 4M_1 M_3)g` ........ (10) SIMILARLY from (6) and (9) `a_3=(3M_1M_2 - M_2 M_3 - 4M_1 M_3)/( M_1 M_2 + M_2 M_3 -4M_1 M_3)` ........ (11) |
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| 14119. |
Calculate the acceleration of the block B of the figure Fig. 5.23, assuming surfaces and pulleys to be smooth. |
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| 14121. |
Define one metre. |
| Answer» Solution :One meter is defined as the LENGTH of the PATH travelled by LIGHT in VACUUM in `1//299,792,458` of a second. | |
| 14122. |
Discuss coefficient of viscosity in the contex of stress and strain . |
Answer» Solution :In figure laminar flow of LIQUID is shown. A layer of liquid sandwiched between two parallel glass plates in which the lower plate is fixed and the upper one is moving to the right with velocity v Liquid layers are shown between the glass plates . Upper plate move with velocity v hence layer contact with it also move with velocity v. On account of this motion , a portion of liquid, which at some instant has the shape of ABCD, take the shape of AEFD after short INTERVAL of time `Deltat`. During time ,the liquid has undergone a shear strain of `(Deltax)/(l)` .The strain in a flowing fluid increases with time continuously. In this case , the stress depend on the rate change of strain or stress depend on the rate change of strain or strain rate , that is `(Deltax)/(lDeltat) or (v)/(l)` instead of strain itself. (Here `(Deltax)/(Deltat)` =velocityv) Here SHEARING stress is `(F)/(A)` . Where A is the area of contact surface and F the viscous force F in the direction of tangent. For fluid , coefficient of viscosity `=eta=("Shearing stress")/("Shear strain")` `thereforeeta=((F//A))/((v//l))=(Fl)/(vA)`....(1) OR `F=etaA((v)/(l))` ....(2) The SI unit ofviscosity is poiseiulle (Pl). 1 poiseiulle `1Nsm^(-2)=P_(a)*s` CGS unit of coefficient of viscosity is poiseiulle 1 poiseiull =1 dyne S `cm^(-2)` 1 poiseiulle =10 pois The dimensions of viscosity are `[M^(1)L^(-1)T^(-1)]`Generally thin liquids like water , alcohol are less viscous then THICK liquids lke coal tar , blood, glycerine etc. The viscosity of liquids decreases with temperature while it increases in the case of gases. 
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| 14123. |
Two simple pendulums of lengths 0.5m and 2.0m respectively are given small linear displacement in one direction at the same time. They will again be in phase when the pendulum of shorter length has completed…………..oscillations. |
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Answer» 5 |
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| 14124. |
A 0.5 kg ball is suspended to a string of length 1m and is rotated in a vertical plane by holding one end of the string. If it is making 30 rpm, the maximum tension in the string is (pi^(2) = 10) |
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Answer» 19.9 N |
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| 14125. |
consider the cyclic process ABCA on a sample of 2.0 mol of an ideal gas as shown in the fig. temperature of the gas at A and B are 300 K and 500 k respectively. A total of 1200 J heat is withdrawn from the sample in the process. Find the work done by the gas in part BC. takeR=8.3 JK^(-1) mol^(-1) |
| Answer» Solution :`W _(BC) = W _(AB) + W _(n et) = 4 P ( V _(B) - V _(A)) + W _(n et) = 4 R ( 503 - 303) + W _(n et) = 6648 + 2000 = 8648 J` | |
| 14126. |
What is the value of acceleration due to gravity on the surface of earth if the entire earth was made of gold ? (Radius of earth = 6400 km , Density of gold = 19.3 xx 10^(3) kg//m^3 and G = 6.67 xx 10^(-11) MKS) |
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Answer» Solution :`IMPLIES` Here, `R_e = 6,400 km . = 64 xx 10^(5) m` `rho = 19.3 xx 10^(3) kg./m^3` `G = 6.67 xx 10^(-11) Nm^(2)//kg^(2)` Mass of EARTH Me = `4/3piRe_^3 rho` `implies g = (GM_e)/(R_e)` `implies g = (G)/(R_e^2) (4/3piRe^2rho)=4/3 piGrhoRe` `:. g = (4 xx 3.14 xx 6.67 xx 10^(-11) xx 19.3 xx 10^(3) xx 64 xx 10^5)/3` `:. g = (12.56 xx 6.6xx 19.3 xx 64 xx 10^(-3))/3` ` :. g = 34.49 m//s^2` |
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| 14127. |
A convex lens and convex mirror are separated by distance d as shown in the figure. What should be the value of d so that image is formed on the object itself. |
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Answer» Solution :For LENS `1/v-1/(-15)=1/10 implies v=+30cm` Case I: IF d=30, the object for MIRROR will be at pole and its image will be FORMED there itself.  Case II: IF the rays stike the mirror normally, they will retrace and the image will be formed on the object itself `THEREFORE d=30-20=10cm` 
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| 14128. |
A stone is thrown horizontally with velocity g ms^(-1) from the top of a towerof height g metre. The velocity with which it hits the ground is (in ms^(-1)) |
| Answer» Answer :C | |
| 14129. |
A box of mass 8 kg is placed on a rough inclined plane of inclination theta. It downward motion can be prevented by applying an upward pull F and it can be made to slide upwards by applying a force 2F. The coefficient of friction between the box and the inclined plane is |
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Answer» `(1)/(3)tan THETA` |
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| 14130. |
Two balls are projected simultaneously with the same speed from the top of a tower - one upwards and the other downwards. If they reach the ground in 6s and 2s, the height of the tower is |
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Answer» 120m |
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| 14131. |
A train of mass 800metric tons is moving with a velocity of 72 KMPH. If the resistive force acting on the train is 10N per metric ton, the power of the engine is |
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Answer» 16KW |
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| 14132. |
If x times momentum is work, then the dimensions of x are |
| Answer» Answer :A | |
| 14133. |
A lorry and a car of mass ratio 4:1 are moving with KE in the ratio 3 : 2 on a horizontal road. Now brakes are applied and braking forces produced are in the ratio 1 : 2. Then the ratio of stopping timings of lorry and car is |
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Answer» `1:1` |
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| 14134. |
The bulk modules of a spherical is 'B'. If it is subjected to uniform pressure 'p' the fractional decrease in radius is: |
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Answer» <P>`(P)/(3B)` |
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| 14135. |
A boy mass m runs on ice with velocity v_() and steps on the end of a plank of length l and mass M which is perpendicular to his path. (a). Describe quatitatively the motion of the system after the boy is on the plank. Neglect friction with the ice. (b). One point on the plank is at rest immediately after the collision. Where is it? |
Answer» Solution : C is the COM of `(M+m)` `BC=((M)/(M+m))(l)/(2)` LTBR and `OC=((m)/(M+m))((l)/(2))` from conservation of linear momentum, `(M+m)v=mv_(0)` or `v=((m)/(M+m))v_(0)`.(i) From conservation of angular momentum about point C we have `mv_(0)(BC)=Iomega` or `(mMv_(0)l)/(2(M+m))=[m((M)/(M+m))^(2)((l)/(4))^(2)+(Ml^(2))/(12)+M((m)/(M+m))^(2)((l^(2))/(4))]omega` putting `(mv_(0))/(M+m)=v` From Eq (i) we have `(v)/(omega)=(l)/(6)[(4m+M)/(M+m)]` Now a point say P at a DISTANCE `x=(v)/(omega)`, from C (tawards O) will be at rest Hence distance of point P from by at B will be `BP=BC+x` `=((M)/(M+m))((l)/(2))+(l)/(6)[(4m+M)/(M+m)]` `=(2l)/(3)` |
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| 14136. |
The normal density of gold is rho. It.s bulk modulus is K. Find the increase in density of a piece of gold when a pressure Pis applied uniformly from all sides? |
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Answer» <P> Solution :`K=(P)/((dV)/(V)) rArr (dV)/(V)=(P)/(K) and RHO =(M)/(V), rho +Delta rho=(M)/(V-DeltaV) rArr rho=(M)/(V-DeltaV)-(M)/(V)``Delta rho=(M)/(V)[(1)/(1-(DeltaB)/(V))-1] rArr Deltarho=(M)/(V)[(1)/(1-(P)/(K))-1] rArr Delta rho=(M)/(V)[(K)/(K-p)-1] rArr Delta rho =(M)/(V) [(K-K+P)/(K-P)]` `Delta rho=(rho P)/(K-P)` |
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| 14137. |
Three moles of a diatomic gas is taken at temperature T. Its volume is varied according to the law V=alpha T ^(-2) where alphais a positive constant during the process. Final temperature of the gas is found to be 2T. Find the heat supplied to the gas. |
| Answer» SOLUTION :`3/2 RT ` | |
| 14138. |
There are three iron rods and two brass rods in a compensated pendulum. Coefficients of linear expansion of iron and brass are 12xx10^(-6@)C^(-1)and19xx10^(-6@)C^(-1) respectively. If the average length of each iron rod is 50 cm, then what is the average length of each brass rod? |
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| 14139. |
The velocity of liquid flowing through a tube at certain distance from the axis of tube |
Answer» Solution :Construction:It consists of two wider tubes A and A' (with cross sectional area A) connected by a narrow tube B (with cross sectional area a). A MANOMETER in the form of U-tube is also attached between the wide and narrow tubes as shown in Figure. The manometer contains a liquid of density `'rho_(m)'`. Theory:Let `P_(1)` be the pressure of the fluid at the wider region of the tube A. Let us ASSUME that the fluid of density `'rho'` flows from the PIPE with speed `'v_(1)'` and into the narrow region, its speed increases to `'v_(2)'`. According to the Bernoulli's equation, this increases in speed is accompanied by a DECREASE in the fluid pressure `P_(2)` at the narrow region of the tube B. Hence, the pressure difference between the tubes A and B is noted by measuring the height difference `(DeltaP=P_(1)-P_(2))` between the surfaces of the manometer liquid. From the equation of continuity, we can say that `Av_(1)=av_(2)` which means that `v_(2)=(A)/(a)v_(1)` Using Bernoulli's equation `P_(1)+rho(v_(1)^(2))/(2)=P_(2)+rho(v_(2)^(2))/(2)=P_(2)+rho(1)/(2)((A)/(a)v_(1))^(2)` From the above equation, the pressure difference `DeltaP=P_(1)-P_(2)=rho(v_(1)^(2))/(2)((A^(2)-a^(2)))/(a^2)` Thus, the speed of flow of fluid at the wide end of the tube A `v_(1)^(2)=(2(DeltaP)a^2)/(rho(a^(2)-A^(2)))` `rArrv_(1)=sqrt((2(DeltaP)a^(2))/(rho(a^(2)-A^(2))))` |
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| 14140. |
A body slides down an inclined plane of inclination theta. While sliding downwards the coefficient of friction is directly proportional to the displacement. The body slides down the plane with a |
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Answer» constant acceleration gsin `theta` |
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| 14141. |
Match the followings : |
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| 14142. |
A man of 50 kg is standing on the school play ground at Trichy. The latitude of Trichy is 108^(@). Calculate the time (in hour ) to complete one rotation (one day) of the earth with the new angular speed. |
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Answer» Solution :Time TAKEN to complete one rotation of the earth (T)=? New ANGULAR speed `omega_(" new ") = 1.261xx 10^(-3)` rad/s `omega = (2pi)/(T) implies T=(2pi)/(omega) = (6.28)/(1.261xx10^(-3))` `=4.985 10^(3) SEC` `T= (4.985 10^(3))/(3600)= 1.4h`. |
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| 14143. |
A liquid column of height 80 cm at 0^(@)C balances the same liquid of height 80.4 cm at 100^(@)C. gamma_(R) is |
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Answer» `4 xx 10^(-5) //°C` |
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| 14144. |
A vihicle of mass 500 kg is moving with a velocity of 15ms^(-1) . It is brought to rest by a retarding force . Find the distance moved by the vehicle before coming to rest , if the sliding friction between the tyres and the road is 3000N. |
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| 14145. |
A gas at 10^(@)C temperature and 1.013xx10^(5) Pa pressure is compressed adiabatically to half of its volume. If the ratio of specific heat of the gas is 1.4, What is its final temperature?[2^(0.4)=1.32] |
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Answer» `101^(@)C` |
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| 14146. |
State the changes in time period of a pendulum, when (i) the pendulum is taken to a mountain top from the earth's surface (ii) the pendulum is set on the floor of a mine (iii) diameter of the bob of the pendulum is increased (iv) keeping the radius of the bob unchanged, its mass is increased. Give reasons for your answer. |
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Answer» Solution :TIME period of a simple pendulum `T=2pisqrt(L/g)`. (i) Taking the pendulum from the earth.s surface to a mountain top, the VALUE of g decreases. Hence, time period increases. (II) On the FLOOR of a mine, value of g decreases. Hecne, time period of a pendulum increases. (iii) On increasing the diameter of the bob, its effective length increases. Hence, time period increases. (iv) Time period is independent of the mass of the pendulum bob. Hence, even if mass is increased, keeping diameter same, time period REMAINS unchanged. |
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| 14147. |
A body of 500 g is tied to one end of a string 2 m long and is revolved in a horizontal circle. If the string breaks under a load of 4.84 kgf, find the maximum number of revolutions per minute the body can make without breaking the string. (g = 9.8 mcdot s^(-2) ) |
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| 14148. |
When you take 500 obseravations instead of 100 observations of a measurement, by what factor is probable error reduced ? |
| Answer» SOLUTION :The PROBABLE ERROR REDUCES to `1//5 TH.` | |
| 14149. |
If x, v & a denote the displacement , the velocity & acceleration of a particle executing simple harmonic motion of time period T, then which of the following does not change with time . |
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Answer» `a^(2)T^(2)+4PI^(2)V^(2)` |
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| 14150. |
A sphere falls in air with a terminal velocity of 50 m/s. If the same sphere is allowed to fall in vacuum what is its terminal velocity? |
| Answer» SOLUTION :There will be no TERMINAL VELOCITY | |