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.
| 40001. |
The moment of inertia of a fly wheel making 300 revolutions per minute is 0.3 kgm^(2) . Find torque required to bring it to rest in 20 s. |
|
Answer» Solution :`n_(1)`=300 RPM=`(300)/(60)=5rps,1=0.3 kgm^(2)` `omega_(1)=2pin_(1)=2pixx5=10pi,omega_(2)=2pin_(2)=0` `THEREFORE tau=(0.3xx(0-10pi))/(20)=(-0.3xx10pi)/(20)=-0.15 pi=-0.471 NM` |
|
| 40002. |
Explain the use of air bags in cars during accidents. |
| Answer» Solution :Carsaredesignedwith air BAGS in suchwaythatwhenthe carmeetswithan accidentthemomentumof thepassengerswillreduceslowly. Sothatthe average FORCEACTING ONTHEM WILLBE smaller. | |
| 40003. |
To maintain a rotor at a uniform angularspeed of 200 rad s^(-1) , an engine needs to transmit a torque of 180 N m. The power required by the engine. Assume that the engine is 100% efficient. |
| Answer» ANSWER :A | |
| 40004. |
The dimensional formula of Areal velocity is |
|
Answer» `M^(@) L^(2) T` |
|
| 40005. |
The escape speed from the surface of the Earth is given by________ |
|
Answer» `SQRT(2gR_E)` |
|
| 40006. |
When air plane fly at higher altitude it is not right to keep a fountain pen in a pocket. |
| Answer» Solution :As the ALTITUDE increases pressure DECREASES . The ink in the refil of pen is filled at atmospheric pressure . Whenair plane fly up the atmospheric pressure on the ink decreases and hence ink COMES out on cloth .Hence , when the air plane fly up , it is not rightto KEEPA fountain pen in a pocket . | |
| 40007. |
Terminal velocity of a copper ball of radius 2 mm through a tank of oil at 20^(@)C is 6.0 cm/s. Compare coefficient of viscosity of oil. Given p_(cu) = 8.9 xx103 kg//m^(3), rho_(oil) = 1.5 xx 103 kg//m^(3). |
|
Answer» Solution :`nu_(t)=(2)/(9)[(G(sigma-rho)r^(2))/(eta)]` `eta=(2)/(9)[(9.8xx(8.9xx10^(3)-1.5xx10^(3))(2XX10^(3))(2xx10^(-3))^(2))/(6xx10^(-2))]` `=1.08kgm^(-1)s^(-1)` |
|
| 40008. |
The energy required to shift a satellite from orbital radius 'r' to radius 2r is E. What energy will be required to shift the satellite from orbital radius 2r to 3r ? |
|
Answer» E |
|
| 40009. |
Find the acceleration of centre of mass of the blocks of masses m_(1) and m_(2) (m_(1)gtm_(2)) in Atwood's machine : |
|
Answer» Solution :We know from Newton.s laws of MOTION magnitude of acceleration of each BLOCK is `a=((m_(1)-m_(2))/(m_(1)+m_(2)))G`  Now acceleration of their C.M is `a_(cm)=(m_(1)a+m_(2)(-a))/(m_(1)+m_(2))`, `a_(cm)=((m_(1)-m_(2))//(m_(1)+m_(2)))a` `a_(cm)=((m_(1)-m_(2))/(m_(1)+m_(2)))((m_(1)-m_(2))/(m_(1)+m_(2)))g` `:.` ACCLERATION of CENTRE of mass `a_(cm)=((m_(1)-m_(2))/(m_(1)+m_(2)))^(2)g` |
|
| 40011. |
A wall is oftwo layers P and Q eachmade of different material . Boththe layers have same thickness . Thetemperature difference between the wall is 36^(@)C. What is the temperature difference acrossthe layer P ? |
|
Answer» <P> 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)` Let`theta` be THETEMPERATUREAT the junction two layers and `theta_(1), theta_(2)` are the temperatureat the endfaces of P and Q . Given`theta_(1) = theta_(2) = 36^(@)C` under thermal euqilibrium `(Q)/(t) = (k_(1)a_(1)(theta_(1) = theta))/(l_(1)) = (k_(2)A_(2) (theta - theta_(2)))/(l_(2))` |
|
| 40012. |
Derive the relation between momentum and kinetic energy. |
|
Answer» SOLUTION :(i) CONSIDER an OBJECT of mass m moving with a velocity `vec(v)`. Then its linear momentum is `vec(p)=mvec(v)` and its kinetic energy, `KE1/2mv^(2)`. `KE=1/2mv^(2)=1/2M(vec(v).vec(v))"...(1)"` (ii) Multiplying both the numerator and denominator of equation (1) by mass, m `KE=1/2(m^(2)(vec(v).vec(v)))/m` `=1/2((mvec(v)).(mvec(v)))/m[vec(p)=mvec(v)]` `=1/2(vec(p)*vec(p))/m` `=p^(2)/(2m), KE=p^(2)/(2m)` (iii) where `abs(vec(p))` is the magnitude of the momentum. The magnitude of the linear momentum can be OBTAINED by `abs(vec(p))=p=sqrt(2m(KE))` (iv) Note that if kinetic energy and mass are given, only the magnitude of the momentum can be calculated but not the direction of momentum. It is because the kinetic energy and mass are scalars. |
|
| 40013. |
Three uniform spheres each having mass .m. and radius .R. kept in such a way that each two touches the other. The magnitude of the gravitational force on any sphere, due to the other two is |
|
Answer» `SQRT3(GM^2)/(4R^2)` |
|
| 40014. |
An artificial satellite orbiting the earth doesnot falldown, because the earth's attraction |
|
Answer» is balanced by the ATTRACTION of the moon |
|
| 40015. |
Use the assumptions of the previous question. An object weighted by a spring balance at the equator gives the same reading as a reading taken at a depth d below the earth's surface at a pole (d lt lt R). The value of d is |
|
Answer» `(OMEGA^(2) R^(2))/(g)` |
|
| 40016. |
Given below are observations on molar specific heats at room temperature of some common gases. {:("Gas","Molar specific heat "(C_(v))),(,("cal mol"^(1)K^(-1))),("Hydrogen",4.87),("Nitrogen",4.97),("Oxygen",5.02),("Nitric oxide",4.99),("Carbon monoxide",5.01),("Chlorine",6.17):} The measured molar specific heats of these gases are markedly different from those for monatomic gases. Typically, molar specific heat of a monatomie gas is 2.52 cal/mol K. Explain this difference. What can you infer from the somewhat larger than the rest) value for chlorine ? |
| Answer» SOLUTION :The gases are diatomic, and have other degrees of freedom fic, have other MODES of motion) possible besides the translational degrees of freedom. To raise the temperature of the gas by a certain amount, heat is to be SUPPLIED to increase the average energy of all the modes Consequently, molar specific heat of diatomic gases is more than that of monatomie gases. It can be shown that if only rotational modes of motion are CONSIDERED, the molar specific heat of diatomic gases is nearly (5/2) R which agrees with the observations for all the gases LISTED in the table, except chlorine. The higher value of molar specific heat of chlorine indicates that besides rotational modes, vibrational modes are also present in chlorine at room temperature. | |
| 40017. |
Define forced oscillation . Give an example. |
| Answer» Solution :The body executing vibration INITIALLY vibrates with its NATURAL frequency and dur to the presence of external PERIODIC force, the body later vibrates with the fequency of the applied periodic force. Such vibrations are known as FORCED vibrations. | |
| 40018. |
Show that the velocity of travelling wave produced in a string is v= sqrt((T)/(mu)) |
|
Answer» Solution :Velocity of transverse waves in a stretched string: Let us compute the velocity of transverse travelling waves on a string . When a jerk is given at ONE end (left end ) of the rope , the wave pulses move towards right end with a velocity v . This means that the pulses move with a velocity v and respect to an observer who is at rest frame . Suppose an observer also moves with same velocity v in the direction of motion of the wave pulse , then that observer will notice that the wave pulse is stationary and the rope is moving with pulse with the same velocity v . Consider an elemental segment in the string . Let A and B be two points on the string at an instant of time . Let dl and dm be the length and mass of the elemental string , respectively . By definition , linear mass density `mu` is `mu = (dm)/(dl) "" ... (1)` `dm = mu dl "" ... (2)` The elemental string AB has a curvature which looks like an arc of a circle with centre at 0 , radius Rand the arc subtending an angle `theta`at the origin O. The angle `theta`can be written in termsof arc length and radius as `theta= (dl)/(R)` . The centripetal acceleration supplied by the tension in thestring is `a_(cp) = (v^(2))/(R) "" ... (3)` Then , centripetal force can be obtained when mass of the string (dm) is included in equation (3) . `F_(cp) =((dm) v^(2))/(R) "" ... (4)` The centripetal force experienced by elemental string can be calculated by substituting equation (2) in equation ( 4) we get `((dm) v^(2))/(R) = (mu v^(2) dl)/(R) "" ... (5) `  The tension T acts along the tangent of the elemental segment of the string at A and B. Since the arc length is very small, variation in the tension force can be ignored. We can resolve T into horizontal component T cos `((theta)/(2))`and vertical component T sin `((theta)/(2))` The horizontal components at A and Bare equal in magnitude but OPPOSITE in direction, therefore, they cancel each other. Since the elemental arc length AB is taken to be very small, the vertical components at A and B appears to acts vertical towards the centre of the arc and hence, they add up. The net radial force F is `F_(r)=2T sin ((theta)/(2)) "" ... (6)` Since the amplitude of the wave is very small when it is compared with the length of the string , the sine of small angle is approximated as sin `((theta)/(2)) = (theta)/(2)` . Hence equation (6) can be written as `F_(r) = 2 T xx (theta)/(2) = T theta "" .... (7)` But `theta = (dl)/(R)` ,, therefore substituting in equation (7) , we get `F_(r) = T (dl)/(R) "" ... (8)` Applying Newton.s SECOND law to the elemental string in the radial direction, under EQUILIBRIUM , the radial component of the force is equal to the centripetal force. Hence equating equation (5) and equation (8) , we have `T(dl)/(R) = mu v^(2) (dl)/(R)` `v = sqrt((T)/(mu))` measured in `ms^(-1) "" ... (9)` |
|
| 40019. |
A boy standing on a rotating table with heavy spheres in his hands, suddenly brings his hands close to his body. The angular velocity of the table |
|
Answer» REMAINS unchanged |
|
| 40020. |
Explain least count and least count error. Write a note on least count error. |
|
Answer» SOLUTION :The smallest value that can be measured by the measuring instrument is called its least COUNT. All the readings or measured values are good only up to this value. Error associated with resolution of the instrument is called least count error. Least count of vernier is 0.01 cm and least count of spherometer is 0.001 cm. Least count error belong to category of random error but within a limited size. It OCCUR with both systematic and random error. Least count of meter scale is 1 mm. Using instrument of HIGHER precision, improving experimental technique we can reduce least count error. Repeating the observation several times and taking arithmetic mean of all observation the mean value would be very close to the true value of the measured QUANTITY. |
|
| 40021. |
Bernoulli's equation is based on conservation of ….. |
|
Answer» energy |
|
| 40022. |
Five spheres of masses 1, 2, 3, 4, 5 kg are mocing along a straight line in the same direction with velocities 5m/s, 4m/s, 2m/s, 1m/s respectively. The first sphere collide with the second and both stick together. The compound mass collides with the third sphere and so on the velocity of the compound mass whn all stick together is |
|
Answer» 3/7 m/s |
|
| 40023. |
Which of the following different equations represents a damped harmonic oscillator ? |
|
Answer» `(d^(2)y)/(DT^(2))+y=0` |
|
| 40024. |
The frequency of oscillation is (10/pi) (in Hz) of a particle of mass 0.1kg which executes SHM along X-axis. The kinetic energy is 0.3J and potential energy is 0.2J at position x=0.02m. The potential energy is zero at mean position. Find the amplitude of oscillations (in metre): |
|
Answer» `1/(2sqrt(10))` |
|
| 40025. |
A load of 31.4 kg is suspended from a wire of radius 10^-3 m and density 9 times 10^3 kg//m^3. Calculate the change in temperature of the wire if 75% of the work done is converted into heat. The Young's modulus and the specific heat capacity of the material of the wire are 9.8 times 10^10 N//m^2 and 490 J.kg^-1.K^-1 respectively. |
|
Answer» |
|
| 40026. |
In the above problem the net force acting on the lower support is xMg. The value of 'x' is |
|
Answer» |
|
| 40027. |
Three particles of masses 1 kg, 2 kg and 3 kg are placed at the vertices A, B and C of an equiliateral triangle ABC. If A and B lie at (0, 0) and (1, 0) m, the co - ordinates of their centre of mass are |
|
Answer» `((SQRT(3))/(2)"m and"(7)/(6)m)` |
|
| 40028. |
A spool has the shape shown in figure. Radii of inner and outer cylinders are R and 2 R respectively. Mass of the spool is 3 m and its moment of inertia about the shown axis is 2 m R^(2). Light threads are tightly wrapped on both the cylindrical parts. The spool is placed on a rough surface with two masses m_(1) = m and m_(2) = 2m connected to the strings as shown. The string segment between spool and the pulleys P_(1) and P_(2) are horizontal. The centre of mass of the spool is at its geometrical centre. System is released from rest. (a) What is minimum value of coefficient of friction between the spool and the table so that it does not slip? (b) Find the speed of m_(1) when the spool completes one rotation about its centre. |
|
Answer» |
|
| 40029. |
Which of the following is meaningful? |
|
Answer» VECTOR/Vector |
|
| 40030. |
The ratio of the areas of cross-section of two rods of different materials is 1:2, and the ratio of the thermal conductivities of their materials is 4:3. On keeping equal temperature-difference between the ends of these rods, the rate of conduction of heat are equal. Determine the ratio of the lengths of the rods. |
|
Answer» |
|
| 40031. |
The work done by the Sun's gravitational force on the Earth is: |
|
Answer» ALWAYS zero |
|
| 40032. |
An ideal monoatomic gas is taken round the cycle ABCDA as shown in the P-V diagram. Compute the work done in this process. |
|
Answer» SOLUTION :Let the foot of the perpendiculars drawn from points A,B,C and D on the volume axis be named as `A^(1), B^(1), C^(1) and D^(1)` respectively. AB,BC,CD and DA will be respectively, given by, the AREAS `(+ABB^(1)A^(1)), (+BC C^(1)B^(1)), (-C C^(1)D^(1) D) and (-D D^(1)A^(1)A)`. Here, the work corresponding to expansion is positive, and that corresponding to compression is NEGATIVE. :. Total work done=AREA `(+ABB^(1)A^(1))+"area" (+BC C^(1)B^(1))+ "area" (-C C^(1)D^(1) D) + "area" (-D D^(1)A^(1)A) = "area" (+ABCD)` `=(6V_(0)-3V_(0)) (5P_(0)-3P_(0))` `(3V_(0))(2P_(0))=6P_(0)V_(0)` units . |
|
| 40033. |
When a body is falling freely from some height work done by gravity is |
|
Answer» NEGATIVE |
|