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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.
| 12901. |
Calcuate the length of a narrow tube closed at one end, which will resonate with a tuning fork of frequency 510 Hz if the velocity of sound is 340 ms^(-1). |
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Answer» |
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| 12902. |
The specific heat of air at constant pressure is 1.005 kJ/kg k and the specific heat of air at constant volume is 0.718 kJ/kgK. Find the specific gas constant |
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Answer» 0.287 kj/kg K |
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| 12903. |
The path of the centre of mass of earth-moon system around the sun is |
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Answer» CIRCULAR path |
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| 12904. |
A cylinder of radius r made of a material of thermal conductivity K_1 is surrounded by a cylindrical shell of inner radius r and outer radius 2r made of a material of thermal conductivity K_2. The two ends ofthe combined system are maintained at two different temperatures. There is no loss of heat across the cylindrical surface and the system is in steady state. Show that the effective thermal conductivity of the system is (K_1 + 3K_2 )//4. |
| Answer» Solution :`Q_1 = K_(1) (PI R_2) (T_1 - T_2) //I, Q_2 = [ pi (2R)^(2)- pi R^(2) ] xx (T_1 - T_2 )//L, Q= Q_1 + Q_2 = (K_1 + 3K_2) pi R^(2) (T_1 - T_2) // l "ALSO" Q= K pi(2R)^(2)(T_1 - T_2) // l, K_1 + 3K_2 = 4K, K= K_1 + 3K_2//4` | |
| 12905. |
A uniform square plate in xy plane has a small piece Q of a irregular shape removed and glued to the centre of the plate leaving a hole behind. The moment of inertia about the z-axis then |
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Answer» increased |
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| 12906. |
In the arrangement shown in figure, there is a friction force between the blocks of masses m and 2m. The mass of the suspended block is m. The block of mass m is stationary with respect to block of mass 2m. The minimum value of coefficient of friction between m and 2m is |
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Answer» `(1)/(2)` `a=("Net pulling FORCE")/("Total mass")` `therefore"" mug=(mg)/(4m)"or"mu=(1)/(4)` |
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| 12907. |
The value of .g. at a particular point on the surface of the earth is 9.8 ms^(2). Suppose the earth suddenly shrinks uniformly to half its present size without losing any mass, the value of g at the same point will now be (assuming that the distance of the point from the centre of the earth does not change ) |
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Answer» `4.9ms^(-2)` |
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| 12908. |
Whichof the followingis the dimension of coefficientof friction |
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Answer» `MLT^(-2)` |
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| 12909. |
The driver of a three -wheeler moving with a speed of 36 km/ h sees a child standing in the middle of the road and brings his vehicle to rest in 4. 0 s just in time to save the child . What is the average retarding force on the vehicle ? The mass of the three wheeler is 400 kg and the mass of the driver is 65 kg . |
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Answer» 1000 N |
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| 12910. |
If the mass of earth is M, radius R and G is universal gravitational constant then the workdone against gravity to move a body of mass 1 kg from the surface of earth to infinity will be |
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Answer» `sqrt((GM)/(2R))` `=(-GM)/(R)` at INFINITY = 0 work = change in potential energy =` (GM)/R` |
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| 12911. |
Two drops of water of the same size fall through air with a uniform velocity of 0.05m. s^-1. If the two drops coalesce to form a single drop, then what will be the terminal velocity of the large drop? |
| Answer» SOLUTION :`0.0794m. S^-1` | |
| 12912. |
Pendulums of two different length are suspended from an elastic chord. If one is set in oscillation will there be any resonance ? Why ? |
| Answer» SOLUTION :There will no be any resonance. Because the length of the PENDULUM is DIFFERENT. Therefore time periods of two different lengthed pendulumsvary. Therefore their FREQUENCIES are different and so they are not in resonance. | |
| 12913. |
A metal plate 5 cm xx 5 cm rests on a layer of castor oil 1 mm thick whose coefficient of viscosity is 1.55 Nsm^(-2) . Find the horizontal foce required to move the plate with a speed of 2 cm/s |
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Answer» Solution :`"" ` Area = ` A = 25 cm ^(2)= 25 XX 10 ^(-4)m^(2)` `"" ` THICKNESS = ` dx = 1 mm = 10^(-3)m ` COEFFICIENT of velocity ` = eta = 1.55 Ns m^(-2)` ` "" ` Force= ? ` "" dv = 2 xx 10 ^(-2)m//s` ` "" F = eta( ADV)/( dx)` ` "" = 1.55 xx 25 xx 10 ^(-4) xx ( 2xx10 ^(-2))/( 1 xx 10 ^(-3))` ` = 0.0775 ` N |
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| 12914. |
The satellite of mass m revolving in a circular orbit of radius r around the earth has kinetic energy E. Then its angular momentum will be: |
| Answer» Answer :C | |
| 12915. |
(A) For collision, momentum transfer takes place.(B) In ant collision, energy transfer takes place.(C ) Impulsive forces are involved in collisions. |
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Answer» A and C are true |
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| 12916. |
We are standing in a stationary bus. When bus suddenly start why we are thrown in backward direction ? |
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Answer» Solution :Whenwe are standinginstationarybus ourfeetare intouchwiththe FLOOR. Frictionalforceis betweenfeetand floorof thebus. Whenbussuddenlystartdueto frictionourfeetstartmovingin DIRECTIONOF buswhereasrest ofthe bodytryto retainpositiondue toinertia. hencewe arethrownin backwarddirection inoppositedirection ofof buswhereasrestofthe bodytryto retainpositiondue toinertia hencewe arethrowninbackwarddirection inoppsitedirectionof ACCELERATIONOF bus. |
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| 12918. |
Discuss the explosion of projectile. |
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Answer» SOLUTION :Here is figure a projectile FOLLOWING the parabolic PATH, explodes into fragment midway in air. The forces leading to the explosion are internal forces, they contribute noting to the motion of the centre of MASS. Total external force, the force of gravity acting on the BODY is the same before and after the explosion. The centre of mass under the influence of the external force continuous therefore, along the same parabolic trajectory as it would have followed if there were no explosion. |
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| 12919. |
A hydrogen -gas -filled balloon ofmass m can rise to a maximum height h above the earth's surface .On the earth's surface, potential energy of the balloon= |
| Answer» Solution :The potential energy of an object of MASS m , raised to a height H is mgh . Here mg=weight of the object =net downward force. But for a balloon ofmass m, the net downward force isnot mg. When it isstationary at a height h, the upward bouyancy balances the weight, the net downward force is ZERO. So the potential energy of the balloon at thatheight=0 , not mgh. this means thatthe total mechanical energy is zero, on the GROUND as well as at the height h. Thus , the law of CONSERVATION of energy is not violated. | |
| 12920. |
Obtain the equation of frequency of oscillations in string tied at both ends. |
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Answer» Solution :Let a string of length L is tied at two ends and it has tension. Its one end is tied at x=0 and another at x=L. For node, displacement sin kL =0 at any point. (Here x =L) `therefore kL = n pi` (where `n = 1,2,3,…,n)` `therefore (2pi L)/(LAMDA ) = n pi ` `therefore L = (n lamda)/(2)""...(1) ` where `n = 1,2,3,...,` `therefore lamda = (2L )/(n) ""...(2) n =1,2,3,...` and `v = lamda v` where v is speed of wave and v is frequency. `therefore (u)/(v) = lamda = (2L)/(n)` `therefore v = (nv)/(2L)""...(3) n = 1,2,3,...,` Equation (2) is for wavelength of stationary wave and equation (3) is for its natural frequency. |
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| 12921. |
The product of torque acting on a body and angular velocity is |
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Answer» Energy |
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| 12922. |
A fighter plane flyinghorizontally at an altitude of 1.5 km with speed 720 km *h^(-1) passes directly over an anti-aircraft gun. At what angle with the vertical shouldthegun be firedfor the shell with muzzle speed 600 m*s^(-1) to hit the plane? At what minimum altitude should the pilot fly the plane to avoid being hit ? (Take g=10 m*s^(-2)). |
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Answer» |
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| 12923. |
A bullet of mass 2 g travelling with a velocity of 500 ms^(-1) is fired into a block of wood of mass 1 kg suspended from a string of 1 m length. If the bullet penetrates the block of wood and comes out with a velocity of 100 ms^(-1), find the vertical height through which the block of wood will rise (assuming the value of g to be 10 ms^(-2)). |
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Answer» Solution :Let the masses of the bullet and the block be .m. and .M. respectively. Let their velocities after the impact be v and V respectively. Let the initial velocity of the bullet be .u.. ACCORDING to the law of conservation of linear MOMENTUM. Mu = mv + MV Here `m = 2xx10^(-3)KG, u=500 ms^(-1), v =100 ms^(-1)` `(2xx10^(-3))xx500=(2xx10^(-3))xx100+(1xxV)` `V=0.8 ms^(-1)`. When the block rises to a height of .h., according to the law of conservation of energy. `(M+m)gh=(1)/(2)(M+m)V^(2)` i.e., `h=(1)/(2)(V^(2))/(G)=((0.8)^(2))/(2xx10)=0.032 m` |
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| 12924. |
The x-component of the resultant of several vectors (a) is equal to the sum of the x-components of the vectors (b) may be smaller than the sum of the magnitudes of the vectors (c) may be greater than the sum of the magnitudes of the vectors (d) may be equal to the sum of the magnitudes of the vectors |
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Answer» only a |
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| 12925. |
Draw a graph showing the variation of g with depth and height from the surface of earth. Assume that the density of the earth is a constant. |
Answer» SOLUTION :
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| 12926. |
IF hat j denotes a unit vector along an incident ray. hat r a unit vector along the refracted ray into a medium of refractive index mu and hat n a unit vector normal to the boundary of the media directed towards the incident medium then the law of refraction can be written as |
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Answer» `HAT I, hat n= (HATR .hatn)` |
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| 12927. |
The pressure is more on the side of |
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Answer» CONCAVE SIDE |
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| 12928. |
The figure shows an insulated cylinder of volume V containing monoatomic gas in both the compartments. The piston is diathermic. Initially the piston is kept fixed, and the system is allowed to acquire a state of thermal equilibrium. If the initial pressures and temperatures are as shown. Calculate the final volume of gas in each compartment. |
| Answer» SOLUTION :`V_("LEFT")=V ((P_1 T_2)/( P_2T_1T_2 ) ) ,V_("RIGHT ")=V((P_2 T_1)/(P_2 T_1 +P_1 T_2))` | |
| 12929. |
Two boys are standing at the eands A and B of a ground where AB = a. They boy at B starts runing in a direction perpendicular to AB with velocity v_(1). The boy at A starts runing simultaneously with velocity v and catches the other in a time t, where t is |
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Answer» `(a)/(SQRT(V^(2)+v_(1)^(2))` |
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| 12930. |
The figure shows an insulated cylinder of volume V containing monoatomic gas in both the compartments. The piston is diathermic. Initially the piston is kept fixed, and the system is allowed to acquire a state of thermal equilibrium. If the initial pressures and temperatures are as shown. Calculate the final temperature and the final pressure. Now the pin which was keeping the piston fixed is removed and the piston set free to move. The piston is allowed to slide slowly, such that a state of mechanical equilibrium is also achieved (T_1 P_2 gt P_1T_2) |
| Answer» SOLUTION :`P_1=(P_1 (P_1 +P_2 )T_2)/(P_1 T_2 +P_2 T_1)` , similarresult for ` P_20` | |
| 12931. |
A circular disc of moment of inertia I_(1) is rotating in a horizontal plane, about its symmetry axis, with a constant angular speed omega_(1). Another disc of moment of inertia I_(2) is placed coaxially on the rotating disc. Initially the second disc has zero angular speed. Eventually both the discs rotate with a constant angular speed omega_(2). The energy lost by the initially rotating disc to friction is ............ |
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Answer» `(1)/(2)(I_(2)^(2))/((I_(1)+I_(2)))omega_(1)^(2)`  According to law of conservation of momentum `therefore I_(1)omega_(1)=(I_(1)+I_(2))omega_(2)` `therefore omega_(2)=(I_(1))/((I_(1)+I_(2)))omega_(1)....(1)` Angular velocity of system decreases when another disc is placed on the first disc. HENCE, `therefore omega_(2)ltomega_(1)` Hence decrease in rotational kinetic energy = (initial rotational kinetic energy) - (final rotational kinetic energy) `=(1)/(2)I_(1)omega_(1)^(2)-(1)/(2)(I_(1)+I_(2))omega_(2)^(2)` `=(1)/(2)I_(1)omega_(1)^(2)-(1)/(2)(I_(1)+I_(2))((I_(1))/(I_(1)+I_(2)).omega_(1))^(2)` putting value of equ. (1) `=(1)/(2)I_(1)omega_(1)^(2)-(1)/(2)(I_(1)+I_(2))(I_(1)^(2))/((I_(1)+I_(2))^(2)).omega_(1)^(2)` `=(1)/(2)omega_(1)^(2)(I_(1))/((I_(1)+I_(2)))(I_(1)+I_(2)-I_(1))` `=(1)/(2)omega_(1)^(2)(I_(1)I_(2))/((I_(1)+I_(2)))` |
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| 12932. |
Two particles of masses m_1 and m_2 are connected to a rigid massless rod of length r to constitute a dumb bell which is free to move in the plane. The moment of inertia of the dumb bell about an axis perpendicular to the plane passing through the centre of mass is |
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Answer» `(m_(1) m_(2)r^2)/(m_(1) + m_(2))`  SUPPOSE C is centre of mass of the DUMB bell , `r_1 , r_2` are DISTANCES of `m_1 , m_2` from C . Therefore , moment of inertia of dumb bell about the GIVEN axis is `I = m_(1) r_(1)^2 + m_(2) r_(2)^2 "" … (i)` Also , `r = r_(1) + r_(2)` and `m_1 r_1 = m_2 r_2 = m_2 ( r- r_1) (m_1 + m_2) r_1 = m_2 r or r_1 = (m_2 r)/(m_1 + m_2)` Similarly , `r_(2) = (m_(1) r)/(m_(1) + m_(2))` From (i) , `I = m_(1) ((m_(2) r)/(m_(1) + m_(2)))^(2) + m_(2) ((m_(1) r)/(m_(1) + m_(2)))^(2)` ` I = (m_(1) m_(2) r^(2))/(m_(1) + m_(2))` |
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| 12933. |
Ball A is dropped from the top of a building and at the same instant that a ball B is thrown vertically upward from the ground. When the balls collide, they are moving in opposite directions and the speed of A is twice the speed of B. At what fraction of the height of the building did the collision occur ? |
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Answer» Solution :Given `V_(A)=2V_(B) rArr gt =2(u-gt) rArr t=(2U)/(3g)` Displacement of A is `S_(A)=1/2 gt^(2) =(2u^(2))/(9G)` Displacement of B is `s_B=ut-1/2 gt^(2)=(4U^(2))/(9g)` Now fraction of hightt of the BUILDING where CONDITION occur is `h_B/H=((4u^(2))/(9g))/((4u^(2))/(9g)+(2u^(2))/(9g))=2/3` |
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| 12934. |
A stone is projected from the ground with a velocity of 14 ms^(-1) one second later it clears a wall 2 m high. The angle of projection is (g = 10 ms^(-2)) |
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Answer» `45^(@)` |
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| 12935. |
six forces lying in a plane and forming angles of 60^(@) relative to one another are applied to the centre of a homogeneous sphere with a mass m=6kg. These forces are radially outward and consecutively 1N,2N,3N,4N,5N and 6N. The acceleration of the sphere is |
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Answer» 0 |
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| 12936. |
When……………is conserved,………………may……………. . |
| Answer» Solution :angular momentum , `KE` of rotation , MAY not be conserved. | |
| 12937. |
A motor of power P_(0) is used to deliver water at a certain rate through a given horizontal pipe. To increase the rate of flow of water through the same pipe n times, the power of the motor is increased to P_(1). The ratio of P_(1) to P_(0) is : |
| Answer» Answer :C | |
| 12938. |
A 60 kg man stands on an elevator floor The elevator is going up with constant acceleration of 1.96 m//s^(2). Percentage change in the apparent weight of the person is |
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Answer» 10 |
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| 12939. |
Under which kind of stress, does a body undergo a change in shape without changing its volume? |
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Answer» |
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| 12940. |
A truck of mass 1000 kg accelerates uniformly from rest to a velocity of 15 ms'in 5 seconds. Calculate (i) its acceleration (ii) its gain in KE and (iii) average power of the engine during this period, neglect friction |
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Answer» Solution :(i) `a=(V-u)/t=3m/s^(2)` (II) `"Gain in K.E."=1/2m(v^(2)-u^(2))=1.125xx10^(5)J` (iii) `p=W/t=225000W` |
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| 12941. |
A ball of mass 1 kg is dropped from a height of 3.2 m on smooth inclined plane. The coefficient of restitution for the collision is e= 1//2. The ball's velocity become horizontal after the collision. |
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Answer» the angle `theta=tan^(-1)(1/sqrt2)` `(8costheta)/2=vsintheta` `implies tantheta=1/sqrt(2)` `v=8/sqrt(2)=3sqrt(2)MS^(-1)` `/_\k=1/2xx1[(4sqrt2)^(2)-8^(2)]=-16J` 
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| 12942. |
The radius of circular orbits of two satellites A and B of the earth, are 4R and R, respectively. If the speed of satellite A ais 3V, then the speed of satellite B will be: |
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Answer» 3V/4 |
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| 12943. |
Consider a string fixed at one end . A travelling wave given by the wave equationy = A sin ( omega t - k x) is incident on it . ltbr gt Show that at the fixed end of a string the waves suffers a phase change of pi, i.e., as it travels back as if the wave isinverted. |
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Answer» Solution :Wave equation of incident wave is `y_(1) A sin ( omega t - kx) ( "positive x - direction")` When the ave strikes the fixed end it must be reflected. The wave will now be travelling in the negativex - direction , and so its equation is ` y_(2) A sin ( omega t + kx + PHI ) ( "negative x - direction")` The phase constant has been added to account for any phase change after reflection . Let us take ` x = 0 ` at the fixed end . The behaviour of a wave at particularpositions is governed by appropriate boundary conditions . For fixed ends , the boundary CONDITION is that the end point is a node . The resultant motion due to incident and reflected wave is ` y= y_(1) + y_(2)` `y = A [ sin ( omega t - kx) + sin ( omega t + kx + phi)]`(i) The boundary condition that must be satisfiedby the resultant wave on string is ` y = 0 at x = 0 ` On substituting these values in Eq. (i) , we obtain `sin omega t + sin ( omega t + phi ) = 0 ` `sin omega t = - sin ( omega t + phi)`(ii) Equation (ii) must be satisfied at all times . For the sake of convenience , we take ` omega t = pi //2`. `sin ((pi)/(2) + phi)= -1` `sin theta = -1 ` implies `theta` can have any of the following values. `3 pi//2 , 7 pi //2 , 11 pi//2 ,....` Therefore `phi` can have any of the values `pi , 3 pi , 5 pi` , etc. All these values are PHYSICALLY possible and distinguishable . We CHOOSE the simplest one , so `phi = pi`. |
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| 12944. |
What are the two basic quests in Physics ? |
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Answer» Solution :The TWO basic QUESTS in Physics are : (i) Unification (II) Redcutionism. |
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| 12945. |
The force acting on a particle vecr=(4,6,12)m of a rigid body is vecF=(6,8,10)N. Find the magniude of the torque producing the rotational motion. Axis of rotation is along the unit vector (1)/(sqrt(3))(1,1,1). |
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Answer» SOLUTION :`vectau=vecrxxvecF` The magnitude of the torque with respect to the axis on which the unit VECTOR is `HATN` is `vec(tau_(n))=(vecrxxvecF).hatn` `THEREFORE vecrxxvecF=|(hati,hatj,HATK),(4,6,12),(6,8,10)|` `therefore vecrxxvecF=hati(60-96)-hatj(40-72)+hatk(32-36)` `=-36hati+32hatj-4hatk=(-36hati+32hatj-4hatk)Nm` `therefore` Magnitude of the torque responsible for rotational motion Now, `(vecrxxvecF).hatn=(-36,32,-4).(1)/(sqrt(3))(1,1,1)` `=(1)/(sqrt(3))(-36+32-4)` `therefore vec(tau_(n)).hatn=-(8)/(sqrt(3))Nm` `therefore tau=(8)/(sqrt(3))Nm` |
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| 12946. |
One car while moving towards a cliff, blows a horn with frequency f. If the driver hears an echo of frequency 2f then speed of car in terms of speed of sound is ...... |
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Answer» (A) V/2`  We have `(f _(L))/( v + v _(L)) = (f _(S))/( v + v ._(S))` `therefore (2f)/( v + v _(L)) = (f)/( v - v _(L))` ` therefore 2 (v -v _(L)) = v + v _(L)` `therefore 2v -2v _(L) =v + v _(L)` `therefore v = 3v_(L)` `therefore v _(L) = v/3` NOTE : In case of an echo, consider an imaginary source on the side opposite to that of REAL souce with opposite VELOCITY. Here according to sign CONVENTION, velocity in a directon from listner to imaginary source is to be take positive. |
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| 12947. |
An experimenter mesuresquantitites a, b, c and X is calculated from X = (ab^2)/(c^3) if the percentage of error in a,b,c are +- 1%, +-3% and +-2% respectively the p.c. of error in X is |
| Answer» ANSWER :A | |
| 12948. |
A hunter aims his gun and fires directly at a monkey sitting on a tree. The moment the bullet leaves the barrel of the gun, the monkey falls. Will the bullet hit the monkey |
| Answer» Solution :Yes, It is because of the fact that both the MONKEY and the BULLET FALLS through the same HEIGHT due to gravity | |
| 12949. |
A lump of iron of mass 2 kg is heated from 40^(@) to 1000^(@)C . If the heat supplied is 192 kcal, find its thermal capacity. Find also the specific heat of iron. |
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Answer» Solution :HEAT supplied `Delta ` Q = 192 k cal `192 xx 10^(3) xx 4.2 `J, MASS m = 2 kg , rise in temperature `Delta t = 1000 - 40 = 960^(@)C`. (a) Thermal capacity of the LUMP of iron = `(Delta Q)/(Delta t) = (192 xx 10^(3) xx 4.2)/(960) = 840 `J/K = (b) Specific heat of iron s = `(Delta Q)/(m Delta t) = (840)/(2)= 420 j kg^(-1) K^(-1)` |
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| 12950. |
An ideal gas is trapped between a mercury column and the closed end of a narrow vertical tube of uniform base containing the column. The upper end of the tube is open to the atmosphere. The atmospheric pressure equals 76 cm of mercury. The lengths of the mercury column and the trapped air column are 20 cm and 43 cm respectively. What will be the length of the air column when the tube is tilted slowly in a vertical plane through an angle of 60^(@)? Assume the temperature to remain constant. |
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Answer» 48 cm |
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