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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.
| 13051. |
A point mass m collides with a disc of mass m and radius R resting on a rough horizontal surface as shown . Its collision is perfectly elastic. Find angular velocity of the disc after pure rolling starts |
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Answer» `((2U)/(3R))` |
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| 13052. |
In a Thermos flask, heat losses due to |
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Answer» CONDUCTION & CONVECTION are MINIMISED by creating vacuum between the two walls of the flask |
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| 13053. |
A stone is dropped from the top of tower. If its velocity at the mid point of height of tower is 10ms^(-1), then the height of a tower is ..... (g = 10 ms^(-2)) |
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Answer» Solution :Suppose initial speed of BALL is `v _(0)` and maximum height is h. Now, for first half distance `d = (h)/(2)` `2AD =v ^(2) -v _(0) ^(2)` `2 xx (-10 ) h/2 = (10) ^(2) - v _(0) ^(2)` `therefore -10 h = 100 -v_(0) ^(2)` `therefore - 10h -100 =- v_(0) ^(2) ""...(1)` Now for maximum height `d = h, v _(0) = v _(0), v =0,` `a = 10 ms ^(-2)` `therefore 2ad =v ^(2) -v_(0) ^(2)` `therefore - 2 xx 10 xx h = 0- v _(0) ^(2) ` `therefore -20h =-v _(0) ^(2) ""...(2)` From equation (1) and (2), `-20 h =- 10h -100` `therefore -10 h =-100` `therefore h = 10M` |
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| 13054. |
The total kinetic energy of a body of mass 10 kg and radius 0.5 m moving with a velocity of 2 m/s without slipping is 32.8 joule.The radius of gyration of the body is |
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Answer» 0.25 m |
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| 13055. |
A tuning fork of frequency 512 Hz makes 4 beats per second with the vibrating string of a piano. The beat frequency decreases to 2 beats per sec. when the tension in the piano string is slightly increased. The frequency of the piano string before increasing the tension was |
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Answer» (A) 510 Hz For piano wire, `f _(B) = 516 OR 508 Hz ` `(because ` Initially beat frequency is 4 beat/second) When tension is increased in piano wire, new beat frequency `f _(A) - f._(B)` becomes 2 Hz. When tension in increased, frequency of piano wire increase. Hence `f ._(B) gt f _(B)` Moreover `f _(A) - f._(B) =2 ` `therefore f _(B) = 508 Hz and f._(B) = 510 Hz` So that `(i) f _(A) - f _(B) = 512 - 508 = 2 Hz` As per the statement, Thus, `f _(B) = 508 Hz` Note Here `f _(B) ne 516 ` because `f ._(B) gt f _(B)` and so `f ._(B) -f_(A) gt 4 Hz` which would be WRONG as it is not so in the statement. |
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| 13056. |
A vessel contains a liquid filled with 1/10th of its volume. Another vessel contains same liquid upto 1/8th of its volume. In both case the volume of empty space remains constant at all temperatures. Then the ratio of coefficient of linear expansions of the two vessels is |
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Answer» `2:5` |
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| 13057. |
Find the number of significant figures in the following numbers. (i) 6729 , (ii) 0.024 ,(iii) 6.0023 ,(iv) 2.520 xx 10^7 , (v) 0.08240 (vi) 4200 , (vii) 4.57xx10^8 , (viii) 91.000 |
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Answer» |
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| 13058. |
A solid sphere of radius .r. is rolling on a horizontal surface. The ratio between the rotational kinetic energy and total energy |
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Answer» `(5)/(7)` |
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| 13059. |
A car travels the first half of a distance between two places a speed of 30//hrand the second half of the distance a 50 km//hr. The averagespeed of the car for the whole joumey is |
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Answer» |
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| 13060. |
The minimum number of non coplanar forces that can keep a particle in equilibrium is |
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Answer» 1 |
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| 13061. |
Apart from linear, circular, vibratory, and rotational motions, other type oof motion like _______and helical motion are also possible |
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Answer» elliptical |
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| 13062. |
A woodrn log of mass 120 kg is floating on still water perpendicular to the shore. A man of mass 80 kg is standing at the centre of mass of the log and he is at a distance of 30 m from the shore. When he walks through a distance of 10 m towards the shore and halts, now his distance from the shore is |
| Answer» ANSWER :B | |
| 13063. |
Assuming the gravity to be in negative z - direction, a force vec(F)=vec(V)xx vec(A) is exerted on a particle in addition to the force of gravity where vec(V) is the velocity of the particle and vec(A) is a constant vector in positive x - direction. What minimum speed a particle of mass m be projected so that it continues to move undeflected with constant velocity ? |
| Answer» SOLUTION :`|VEC(F)|=v` A sin `theta` Here, `theta` is angle between `vec(v)` and `vec(A)` The particle MOVES undeflected if `vec(F)` acts be in positice z - direction. Hence, velocity should be in NEGATIVE y - direction. Or `vec(V_(min))=-(mg)/(A)hat(j)`. | |
| 13064. |
In case of propagation of transverse waves, angle between directions of velocity of particle and velocity of wave is pi rad. |
| Answer» SOLUTION :FALSE. ACTUALLY this ANGLE is `pi/2` RAD. | |
| 13065. |
A brass wire of length 5m and cross section 1mm^(2) is hung from a rigid support, with a brass weight of volume 1000 cm^(3) hanging from the other end. Find the decrease in the length of the wire, when the brass weight is completely immersed in water. (Y_("brass")=10^(11)Nm^(-2), g=9.8ms^(-2), rho_("water")=1g cm^(-3)) |
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Answer» Solution : When a weight is hung in air from the other end of a WIRE, F = Mg. The INCREASE in LENGTH of the wire, e = ? Young.s modulus, `Y = F/A L/e, e = (MgL)/(AY)` When weight hung in a liquid, Weight of the body in the liquid = Mg - `Vrho g ` where V is the volume of the body This is the force, F acting on the wire i.e., `F = mg - V rho g ` Increase in length of the wire, e. ` = ((mg -v rho g)L)/(AY)` which is LESS than the increase in length of the wire when the weight is in air. Decrease in length = e in air - `e^1` in liquid ` = (MgL)/(AY) - ((Mg - V rho g)L)/(AY) = (Vrho gL)/(AY)` Here , ` V = 1000 cm^3 = 1000 xx 10^(-6) m^3` `A = 1mm^2 = 1 xx 10^(-6) m^2 ,Y = 1 xx 10^11 Nm^(-2)` Thedecrease in length ` = (1000 xx 10^(-6) xx 1 xx 10^3 xx 9.8 xx 5)/(1 xx 10^(-6) xx 1 xx 10^11)` ` = 49 xx 10^(-5) m = 0.49 mm` |
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| 13066. |
Find the force required to pull a block of mass 100kg up an incline of 8^(@) at a uniform speed of 5m//s. The coefficient of friction between the block and the plane is 0.3. |
Answer» 
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| 13067. |
A ball of 200g is at one end of a string of length 20cm. It is revolved ina horizontal circle at an angular frequency of 6 rpm. Find (i) the angular velocity, (ii) the linear velocity, (iii) the centripetal acceleration, |
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Answer» Solution :(i) The angular velocity , `OMEGA = (2piN)/(t) = (2pixx6)/(60) = (pi)/(5) = 0.6284 rad s^(-1)`. (ii) Thelinear velocity, `v = romega` `=0.20xx0.6284 = 0.1257m s^(-1)` (iii) The centripetal acceleration, `a_( C) = R omega^(2)` `=0.20xx(0.6284)^(2)=0.0790m s^(-2)`. |
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| 13068. |
A passing aeroplane sometimes causes the rattling of the windows of house. Give reason. |
| Answer» Solution :When the FREQUENCY of sound WAVES from the engine of an AEROPLANE matches with the NATURAL frequency of a WINDOW, reasonance takes place which caused the rattling window. | |
| 13069. |
A diatomic molecule has how many degrees of freedom |
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Answer» 3 |
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| 13070. |
Rain is falling vertically with a speed of 1ms^(-1). A woman rides a bicycle with a speed of 1.732 ms^(-1) in east to west direction. What is the direction in which she should hold her umbrella ? |
Answer» Solution :In FIG. `v_(r)` represents the velocity of rain and `v_(b)`, the velocity of the bicycle, the woman is riding. Both these velocities are with RESPECT to the ground. Since the woman is riding a bicycle, the velocity of rain as experienced by her is the velocity of rain relative to the velocity of the bicycle she is riding. That is `v_(RB) = v_(r) - v_(b)`. This relative velocity vector as shown in Fig. makes an angle `theta` with the vertical. it is given by `tan theta = (v_(b))/(v_(r)) = (SQRT(3))/(1) = sqrt(3) = 60^(@)` Therefore, the woman should hold her umbrella at an angle of about `60^(@)` with the vertical TOWARDS the west. |
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| 13071. |
An ideal gas is taken from the state A (pressure P, volume V) to the state B (pressure P/2 , volume 2V)along a straight line path in the P-V diagram. Select the correct statements(s) from the following. |
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Answer» The work DONE by the gas in the process A to B is negative |
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| 13072. |
A 0.2 kg object at rest is subjected to a force (0.3hati – 0.4hatj)N. What is its velocity vector after 6 sec |
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Answer» `(9hati-12hatj)` |
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| 13073. |
(A): When a bottle of cold carbonated drink is opened, a slight fog forms around the opening. (R ): Adiabatic expansion of the gas causes lowering of temperature and condensation of water vapours. |
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Answer» Both (A) and(R ) are TRUE and (R ) is the CORRECT EXPLANATION of (A) |
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| 13074. |
Four particles each of mass.m. are placed at the corners of a square of side .U. The radius of gyration of the system about an axis normal to the square and passing through its centre is: |
| Answer» ANSWER :A | |
| 13075. |
A perfect gas undergoes the following three separate and distinct process to execute a cycle i)constant volume process during which 80 kJ of heat is supplied to the gas. ii) constant pressure process during which 85 kJ of heat is lost to the surroundings and 20 kJ of work is done on it. iii) adiablatic process which restores the gas back to its initial stage. Evaluate the work done during adiabatic process and the value of internal energy at all the state points if initially its value is 95 kJ. |
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Answer» Solution :This example ILLUSTRATES the application of non-flow energy equation and the `1^(st)` law applied to a cyclic process. Process 1-2: `Q_(1-2)=80 kJ, W_(1-2) = 0` `U_(2)-U_(1) = Q_(1-2)-W_(1-2)=80-0=80 kJ` `U_(2)=80+95=175 kJ`  Process 2-3 : `Q_(2-3)=-85 kJ, W_(2-3)=-20 kJ` `U_(3)-U_(2)=Q_(2-3)-W_(2-3)=-85-(-20)=-65 kJ` `U_(3) = -65+175 = 110 kJ` For the complete CYCLE : `oint delta Q=oint delta W` . That is `Q_(1-2)+Q_(2-3)+Q_(3-1)=W_(1-2)+W_(2-3)+W_(3-1)` `80-85+0=0+(-20)+W_(3-1)` :. Work done during adiabatic process, `W_(3-1)=15 kJ` |
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| 13076. |
Choose the correct pair of factors that are affecting speed of sound in gases. |
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Answer» VOLUME and pressure |
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| 13077. |
An engineer claims to have made an engine delivering 10 kW power with fuel consumption of 1 g/sec . The calorific value of the fuel is 2 kcal/g . Is the claim of the engineer |
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Answer» Valid |
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| 13078. |
The length of wire between the two ends of a sonometer is 105 cm. Where should the two bridges be placed so trhat the fundamental frequencies of the three segments are in the ratio of 1:3:15 ? |
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Answer» SOLUTION :Total length of the WIRE, `L=105 CM` `v_(1):v_(2):v_(3)=1:3:15` Let `L_(1),L_(2) and L_(3)=1:1/3:1/15=15:5:1` Sum of the RATIOS `=15+5+1=21` `therefore L_(1)=15/21**105==75CM, L_(2)=5/21**105=25cm,` `L_(3)=1/21**105=5cm` Hence the bridges should be placed at 75cm and `(75+25)=100cm`from one end. |
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| 13079. |
One end of a thermally insulated metal rod, one metre long and area of cross section 13 cm^2 is immersed in boiling water and the other end in ice. It is found that 0.4 Kg of ice melts in one hour, find the coefficinet of thermal conducativity of the material of that rod ? |
| Answer» SOLUTION :`373.33 WM^(-1)K^(-1)` | |
| 13080. |
A copper wire of length 1m is stretched by 1cm. Find the strain on the wire |
| Answer» SOLUTION :The STRAIN `= (e)/(L) = (1 xx 10^(-2))/(1)= 0.01` | |
| 13081. |
The kinetic energy of rotation K depends on the angular momentum J and moment of inertia I. Find the expression for kinetic energy. |
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Answer» Solution :LET `X prop J^(a)I^(b)` then, `K=CJ^(a)I^(b)"….(1)"` Writing dimensions of both sides, We get, `[ML^(2)T^(-2)]=[ML^(2)T^(-1)]^(a).[ML^(2)]^(b)` `[ML^(2)T^(-2)]=[M^(a+b)L^(a+b)L^(2a+2b)T^(-a)]` Comparing powers of T, we get `-a=-2 or a=2` Comparing powers of M, we get `a+b=1 or 2+b=1 or b=-1` Putting these values of 'a' and 'b' in equation (1) We get, `K=(CJ^(2))/(I)` The value of CONSTANT C cannot be found. |
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| 13082. |
shows a man of mass 60 kg standing on a light weighing machine kept in a cabin of mass 40 kg The cabin is hanging from a pulley fixed to the ceiling through a light rope, the other end of which is held by the man himself if the man manages to keep the cabin at rest, what is the weight shown by the machine? What force should he exert on the rope to get his correct weight on the machine? |
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Answer» |
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| 13083. |
Find the angle between the vectors (-2i + 2j) and 3i |
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Answer» `60^(@)` |
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| 13084. |
The number of significant figures in 0.0160kg is |
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Answer» 2 |
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| 13085. |
When an object is said to be in oscillation ? State examples. |
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Answer» SOLUTION :When an object moves back and FORTH repeatedly for some duration of time its motion is said to be in oscillation. Eg: Our HEART beat, swinging motion of the wings of an INSECT, pendulum CLOCK. |
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| 13086. |
A 0.4kg body is rotated with a constant angular speed of 2 rps in a verticle circle of radius 1.2m with the help of a string. The tension in the string when it is at the highest point is (g= 10 ms^(-2) and pi^(2)= 10) |
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Answer» 79.8N |
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| 13087. |
A particle moves along a horizontal circle with constant speed. If 'a' is its acceleration and 'E' is its kinetic energy (A) a is constant (B) E is constant (C ) a is variable (D ) E is variable |
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Answer» A and B are correct |
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| 13088. |
Two wedights w_(1) and w_(2) are suspended to the ends of a string passing over a smooth pulley. If the pulley is pulled up with acceleration g then find the tension in the string |
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Answer» Solution :Tension = `(2m_(1)m_(2))/(m_(1)+m_(2))(g+a)=(2(w_(1))/(g).(w_(2))/(g))/((w_(1))/(g)+(w_(2))/(g))(g+g)=(4w_(1)w_(2))/(w_(1)+w_(2))` |
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| 13089. |
Consider the following two equations A) L=Iomega (B) (dL)/(dt)=tau |
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Answer» both A and B are true |
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| 13090. |
A uniform rod of mass 4m and length 2r can rotate in horizontal plane about a vertical axis passing through its centre O. Two small balls each of mass m are attached to its ends. A fixed gun fires identical balls with speed v in horizontal direction. The firing is being done at suitable intervals so that the fired balls either hit the ballat end A or B while moving in the direction of velocity of A or B. All collisions are elastic. (i) Initial angular velocity of the rod is zero and its angular velocity after n^(th) collision is omega_(n) Write omega_(n + 1) in terms of omega_(n) (ii) Solve the above equation to get omega_(n) (iii) Find the limiting value of omega. |
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Answer» (II) `omega_(n) = (v)/(R) [1-((7)/(13))^(n-1)]` (iii) `(v)/(r)` |
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| 13091. |
A block of mass 25 kg is raised by a 50 kg man in two different ways as shown in Fig.What is the action on the floor by the man in the two cases ?If the floor yields to a normal force of 700 N, which mode should the man adopt to lift the block without the floor yielding ? |
| Answer» Solution :Consider the FORCES on the man in equilibrium : his WEIGHT, force due to the rope and NORMALFORCE due to the floor. (a)750 N (b)250 N, MODE (b)should be adopted. | |
| 13092. |
A sphere of mass M_(1), moving with a velocity v_(0) collides head on with a stationary sphere of mass M_(2). The collision is elastic. V_(1) and V_(2) are respectively their velocities immediately after collision.{:(,"List - I",,"List - II"),((a),M_(1)=M_(2),(e ),V_(1)=-V_(0)","V_(2)=0),((b),M_(1)lt lt M_(2),(f),V_(1)=0","V_(2)=V_(0)),((c ),M_(1)gt gt M_(2),(g),V_(0)lt V_(2)lt 2V_(0)),((d),2M_(1)gt M_(1)+M_(2),(h),V_(1)=V_(0)","V_(2)=2V_(0)):} |
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Answer» a-g, b-e, C-h, d-f |
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| 13093. |
What is phase difference between the displacement and acceleration of a particle executing S.H.M ? |
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Answer» SOLUTION :Acceleration `PROP` displacement `:.a=-omega^(2)x` `:.` Phase difference `=180^(@)`or `pi` RADIAN. |
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| 13094. |
If oversettoA+oversettoB=oversettoA-oversettoB, what can you say about B? Give two physical examples of such vectors. |
| Answer» Solution :`oversettoA+oversettoB=oversettoA-oversettoB,`when is a NULL VECTOR. The acceleration vector of an object moving with uniform VELOCITY, the velocity vector of a STATIONARY object are examples of a null vector. | |
| 13095. |
body against friction is mgS 12. |
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Answer» Work DONE is indepenent of the path |
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| 13096. |
A small bucket of mass M kg is attached to a lone inextensible cord of length L m. The bucket is released from rest when the cord is in a horizontal position. At its lowest position, the bucket scoops up m kg of water and swings up to a height h. The height h in meter is :- |
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Answer» `((M)/(M + m))^(2) L` COLM : `Msqrt(2gL) = (M + m)V` ……(i) COLM : `(1)/(2) (M + m)v^(2) = (M + m)gh` …..(ii) `rArr v = SQRT(2gh) = (Msqrt2gL)/((M + m)) rArr H = ((M)/(m + M))^(2) L` |
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| 13097. |
A satellite takes 1/8years to move round the earth in 8 its permissible orbit of radius R. The period when it revolves round the earth in an orbit of radius .2R. is |
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Answer» `1/(2sqrt(2))` YEARS |
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| 13098. |
A wire of length 10 m and of cross-section 1 xx 10^(-4)sq.m is fixed at one end and a load of 1200 kg is attached at the other end. Then the extension produced is (Y=9.8xx10^(10)N//m^(2)) |
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Answer» `12xx10^(-3)m` |
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| 13099. |
State the condition in which gravitational potential energy is said to be(i) Negative, (ii) Positive. |
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Answer» SOLUTION :`W=U(r)-U(r')` (i) If `rltr'` Since gravitational force is attractive, since a BODY of mass `m_(2)` is attracted by another body is mass `m_(1)`. Then `m_(2)` can move from r to r' without any external work. Here work is done by the SYSTEM spending its internal energy and hence the work done is SAID to be negative. (ii) If `rgtr'` Work has to be done-against gravity to move the object from r' to r. Therefore work is done on the body by external force and hence work done is positive. |
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| 13100. |
Two equal drops of water are falling through air with a steady volocity of 10 cm//s. If the drops recombine to form a single drop what would be their terminal velocity ? |
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Answer» SOLUTION :` V prop r^(2), v_1 //v_2= r^(2) //R^(2) , 2 XX (4//3) pi r ^(3)= (4//3)pi R^(3),( 8//R)^(2)= 1//(2)^(2//3)` ` V_2 =v_1 (R//r)^(2)= 10 xx 2 ^(2//3)= 15. 9cm //s = 0.16 m//s ` |
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