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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.
| 14951. |
Four spheres a,b,c,d of same material and equal volume are dropped into four liquids of densities d_(1),d_(2),d_(3) and d_(4)(d_(1)gtd_(2)gtd_(3)gtd_(4)). The ascending order of terminal ve-locities (eta is same) |
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Answer» a,b,C,d |
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| 14952. |
The average translational K.E. of O_(2) (molar mass 32 ) molecules at a particular temperature is 0.048 eV. The translational K.E. of theN_(2) (molar mass 28) molecules in eV at the same temperature is ............ . |
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Answer» 0.0015 It is indepent of MOLECULAR MASS. |
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| 14953. |
What is the use of the indicator diagram or PV diagram? |
| Answer» SOLUTION :It is used to find the work DONE by a SYSTEM or on the system. The area enclosed by the P-V graph with the X-axis gives the work done. | |
| 14954. |
The temperature of an object is observed to rise in a period. During this period |
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Answer» Heat must be SUPPLIED to it |
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| 14955. |
A man of mass 80 kg is riding on a small cart of mass 40 kg which is rolling along a level floor at a speed of 2 m/s. He is runing on the cart so that his velocity relative to the cart is 3 m/s in the direction opposite to the motion of cart. What is the speed of the centre of mass of the system. |
| Answer» Answer :D | |
| 14956. |
A shell fired a cannon with speed v m/s at angle theta with horizontal explodes into three pieces of equal masses at the highest point of trajectory. One piece falls down verically while the other retraces its path. What is speed of the third piece ? |
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Answer» SOLUTION :If .m. is mass of shell, at highest point, initial momentum `= m(V cos theta)`, along HORIZONTAL give, momentum of one fragment `= (m)/(3)(-v cos theta)` momentum of second along horizontal = 0 If `.v_(1).` is speed of third fragment, from momentum conservation `-(m)/(3)v cos theta +(m)/(3)v_(1)+0=m(v cos theta)`, `v_(1)=4V cos theta`. |
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| 14957. |
The bob A of a pendulum of length 1 m released from 30^(@) to the vertical hits another bob B of same mass at rest on the table. How high does the bob A rise after the collision and find the velocity of B. |
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Answer» Solution :The BOB of pendulum acquires a velocity `V = sqrt(2gl(1-costheta))`, onreaching the LOWER most POSITION and makes elastic collision with the bob B of same mass.  Hence the bodies simply exchange their velocities after collision. The bob of pendulum does not RISE at all and velocity of B is `V=sqrt(2gl(1-costheta))=1.62ms^(-1)` |
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| 14958. |
The speeds of air-flow on the upper and lower surfaces of a wing of an aeroplane are v_(1) and v_(2) respectively. It A is the cross-sectional area of the wing and 'rho' is the density of air, then the upward lift is |
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Answer» `1//2rhoA(V_(1)-V_(2))` |
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| 14959. |
A particle moves along a circle with acceleration of constant magnitude. Workdone by the resultant force acting on that particle. |
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Answer» is always ZERO |
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| 14960. |
A body starts from rest from a point distance RQ from the centre of the earth. The velocity acquired by the body when it reaches the surface of the earth will be (R —> Radius of the earth) (R_(0) gt R) |
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Answer» `2GM(1/R-1/R_(0))` |
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| 14961. |
A uniform wire of length 12 m and mass 6 kg is suspended from a rigid support. A mass of 2 kg is a attached to the lower free end. A transverse wave of length 0 . 0 6 m is generated at the lower end of the wire . What is its wavelength when the wave reaches the upper end ? |
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Answer» Solution :Velocity of TRANSVERSE vibration in a stretchedwire, ` V= sqrt((T)/(m))` , where T = tension in the wire and m = mass per unit length = constant, for the uniform wire. If`V_(1) and V_(2)` are the velocities at the lower and upper ends, respectively, `(V_(1))/(V_(2)) = sqrt((T_(1))/(T_(2)))` Here, `T_(1)=`tension at the lower end = weight of the suspended mass ` 2 XX 9. 8 ` N , `T_(2)=`tension at the upper end = weight of the wire and the suspended mass ` = ( 2 + 6) xx 9 . 8 = 8 xx 9 . 8 `N Theentire wire will emit a sound of a FIXED frequency, so the wavelength `lambda` is proportional to the wave velocity V . `:. lambda_(2) = (V_(1))/(V_(2)) = sqrt((T_(1))/(T_(2)))` or,` lambda_(2) = lambda_(1) sqrt((T_(2))/(T_(1))) = 0 . 06 xx sqrt((8 xx 9 . 8)/(2 xx 9 . 8)) = 0 . 12 ` m . |
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| 14962. |
Calculate the work done in blowing a soap bubble from radius 2cm to 5cm. Surface tension of water =60 "dyne" cm^(-1). |
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| 14963. |
One Astronomical unit is......... |
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Answer» `3.153 XX 10^(7)` m |
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| 14964. |
A force of 50 Nact on the object of mass 20 kg calculate the acceleration of the object in x and y direction |
Answer» Solution :  Given F = 50N and m= 20 KG (i) component of force along x-direction `F_x = F COS theta` `= 50 xx cos 30^@ = 43.30 N` ` therefore a_x = (F_x)/(m)= (43.30)/(20) = 2.165 MS^(-2)` (ii) Component of force along y-direction ` F_y = F sin theta = 50 xx sin 30^@ = 25N` ` therefore a_y = (F_y)/(m) = 25/20 = 1.25 ms^(-2)` |
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| 14965. |
The length of an unstretched elastic string is 2a. It is connected to two pegs A and B kept at a distance 2a. To the mid-point C an elastic string of length a is attached. The other end D of this string is pulled at right angles to AB until C is displaced through a distance equal to a/10. Calculate the shift of D assuming the strings have the same moduli of elasticity and the same cross sectional area. |
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| 14966. |
A long capillary tube is dipped in water. The water rises up to certain height. What happens to the water colnmn in the capillary, if the entire arrangement is placed in a freely falling elevator? |
| Answer» Solution :`h=(2scostheta)/(rrhog)`. During free fall, the arrangment is in a STATE of WEIGHTLESSNESS. Effective acceleration G - a = 0. So WATER will rise to fill the entire tube. | |
| 14967. |
Radius of gyration of a disc of mass 5kg about a transverse axis passing through its centre is 14.14cm. Find its radius of gyration about its diameter and hence calculate its moment of inertia about its diameter |
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Answer» SOLUTION :Radius of gyration of a disc about a transverse AXIS PASSING through its centre `K=sqrt((I)/(M))=sqrt((MR^(2))/(2M))=(R )/(sqrt(2))=14.14cm` `( :. I=(MR^(2))/(2))` Radius of the disc. `R=14.14xxsqrt(2)=20cm` Radius of gyration of the disc about its diameter. `K=sqrt((I)/(M))=sqrt((MR^(2))/(4M))=(R )/(2)=(20)/(2)=10cm` Moment of INERTIA about its diameter `=(MR^(2))/(4)=(5(0.2)^(2))/(4)=5.00xx10^(-2)kgm^(2)` |
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| 14968. |
The dimensions of ( mu_0epsilon_0 ) ^ { - \frac { 1 } { 2 } } are |
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Answer» `[ L ^ { 1 / 2 } T ^ { - 1 / 2 } ]` |
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| 14969. |
Explain average velocity and average speed. |
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Answer» Solution :When an object is in motion, its position change with time and with what rate of time it will change its position that can be find in two ways. If we only find time rate then it is speed and if we find time rate with direction then it is velocity Speed: The distance covered by object in unit time is called speed. Average speed: The ratio of total distance covered during travelling to total time taken is called average speed. Its unit is mal and it is a scalar quantity. Hence its value is always positive. Velocity : The displacement covered in unit time is called velocity It is a scalar quantity Average velocity is defined as the change in position or displacement (Axl divided by the time intervals an, in which the displacement occurs: `THEREFORE bar v = (x _(2) - x _(1))/(t _(2) - t _(1)) = (Delta x)/(Delta t)` where` x_(1) and x_(2)`are the positions of the object at time `t_(1) and t_(2)` respectively. The SI unit for velocity is `MS^(-1)`although KMH is used in many everyday applications. For motion in a straight line, the DIRECTIONAL aspect of the vector can be taken care of by .+. and - signs and we do not have to USE the vector notation for velocity. Magnitude of average velocity can be positive, negative or zero. Resultant effect of motion can be known from average velocity Average speed can be greater than or equal to the magnitude of average velocity. For uniform motion at every moment, velocity is equal to average velocity Consider the motion of the car, the portion of the x-1 graph between t = 0s and t = 8s is blown up and shown in figure. 
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| 14970. |
A body of weight 20N is on a horizontal surface, minimum force applied to pull it when applied force makes an angle 60^(@) with horizontal (angle of friction a = 30^(@)) is |
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Answer» 20 N |
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| 14971. |
Molecules in air in the atmosphere are attracted by gravitational force of the earth. Explain why all of them do not fall into the earth just like an apple falling from a tree. |
| Answer» Solution :`IMPLIES` The molecules in AIR are not heavy as COMPARED to apple. So, molecules in air ALSO have motion due to thermal VELOCITY. | |
| 14972. |
If E, M, J and G denote energy, mass, angular momentum and gravitational constant respectively, calculate the dimensions of the quantity(EJ^(2))/(M^(5)G^(2)) . |
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| 14973. |
Describe the anomalous expansion of water: How is it helpful in our lives? |
Answer» Solution :Anomalous expansion of WATER : Liquids expand on heating and contract on cooling at moderate temperatures. But water exhibits an anomalous behavior. It CONTRACTS on heating between `0^(@)C` and `4^(@)C`. The volume of the given amount of water decreases as it is cooled from room temperature, until it reach `4^(@)C`. Below `4^(@)C` the volume increases and so the density decreases. This MEANS that the water has a maximum density at `4^(@)C`.This behavior of water is called anomalous expansion of water. In COLD countries during the winter season, the surface of the lakes will be at lower temperature than the bottom as shown in the Figure, Since the solid water (ice) has lower density than its liquid form, below `4^(@)C`, the FROZEN water will be on the top surface above the liquid water (ice floats). This is due to the anomalous expansion of water. As the water in lakes and ponds freeze only at the top the species living in the lakes will be safe at the bottom. |
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| 14974. |
Whichof the tones will be absent in the note emitted from a string clamped at both ends, when it is struck at a distance of one-third or its length from one end ? |
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Answer» `(##CHY_DMB_PHY_XI_P2_U10_C04_E01_002_A01##)` |
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| 14975. |
Explain tangential or shearing stress and shearing strain |
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Answer» Solution :In figure if two EQUAL and opposite deforming forces are applied to the cross-sectional area of the cylinder and there is relative DISPLACEMENT between the opposite faces of the cylinder. The restoring force per UNIT area developed due to the applied tangential force is known as tangential or shearing stress. As a result of applied tangential force, there is a relative displacement Ax between opposite faces of the cylinder as shown in figure. Due to this, the strain so produced is known as shearing strain and it is defined as the RATIO of relative displacement of the faces to the LENGTH of cylinder L `epsi _(0) = (Delta x)/(L) = tan theta =theta`  Where`theta` is the angular displacement of the cylinder from the vertical. Usually`theta` is very small, tane is nearly equal to angle `theta` A body subjected to shearing stress is shown in figure. When a book is pressed with the hand and pushed horizontally, vertical side makes small angle with horizontal. |
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| 14976. |
At room temperature (25^(0)C) the length of a steel rod is measured using a brass centimeter scale. The measured length is 10 cm. If the scale is calibrated to readto read accurately at temperature 0^(@)C, find the actual length of steel rod at room temperature. |
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Answer» Solution :The brass scale is calibrated to READ accurately at `0^(@)C`, this means at `0^(@)C`, eachdivision of scale has exact 1 cm length. Thus at higher temmperature the division length of scale will be more than 1 cm due to thermal expansion. Thus at higher temperature the scale reading for length measurement is not appropriate and as at higher temperature the division length is more, the length this scale READS will be lesser then the actual length to be measured. For example in this case the length of each division on brass scale at `25^(0)C` is `l_(1" div")=(1cm)[1+alpha_("br")(25-0)]=1+alpha_("br")(25)` It is given that at `25^(0)C` the length of STEEL rod measured is 10 cm. Actually it is not 10 cm, it is 10 divisions on the brass scale. Now we can find the actual length of the steel rod at `25^(@)C` as `l_(25^(0)C)= (10cm)xx l_("1 div")(or) l_("ACT")=10[1+alpha_("br")(25)]` The above EXPRESSION is a general relation usingwhich you can find the actual length of the objects of which lengths are measured by a metallic scale at some temperature other than the graduation temperature of the scale. |
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| 14977. |
The potential energy of a particle in a force field is U = A/(r^2) - B/r where A and B are positive constants and r is the distance of particle from the centre of the field. For stable equilibirum, the stable equilibrium , the distance of the particle is |
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Answer» `B/(2A)` For equilibrium, `(DU)/(dr) = 0` `:. - (2A)/(r^3) + B/(r^2) = 0 "or " (2A)/(r^3) = (B)/(r^2) "or " r = (2A)/(B)` For stable equilibrium, `(dU)/(dr) = 0` `:. - (2A)/(r^3) = B/(r^2) = 0 " or " (2A)/(r^3) = B/(r^2) " or " r = (2A)/(B)` for stable equilibrium, `(d^2U)/(dr^2) > 0` `(d^2 U)/(dr^2) = (6A)/(r^4) - (2B)/(r^3)` `(d^2U)/(dr^2)|_(r = (2A//B)) = (6AB^4)/(16A^4) = (2B^4)/(8A^3) = (B^4)/(8A^3) > 0`. So for stable EQUILIBIRIUM, the DISTANCE of the particle is `(2A)/(B)`. |
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| 14978. |
A through inclined is placed on car moving with a constantvelocity u on horizontal ground . A block of mass M rests on the inclinedplane . Is any work done by force of friction then a dissipation of energy ? |
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Answer» Solution :Yes , the friction , `f =Mg sin THETA "" ( :. "M is at REST")`  Since , BLOCK is not in motion so , no WORK is DONE by the friction no dissipation of energy takes place. |
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| 14979. |
The gravitational intensity at the centre Q of the drumhead defined by a hemispherical shell of uniform mass density has the direction indicated by the arrow (i) a,(ii)b, (iii) c, (iv) zero. |
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Answer» Solution :We know that the GRAVITATIONAL potential is constant at all points inside a spherical shell. Therefore, the gravitational potential GRADIENT at all points inside the spherical shell is zero [i.e., as `V` is constant, `(dV)/(DR) = 0`]. Since, gravitational intensity is equal to negative of the gravitational potential gradient, hence the gravitational intensity is zero at all points inside a HOLLOW spherical shell. This indicates that the gravitational forces acting on a particle at any point inside a spherical shell, will be symmetrtically placed. Therefore. if we remove the upper hemispherical shell, the NET gravitational force acting on the particle at the centre `Q` or at some other point `P` will be acting downwards which will also be the direction of gravitational intensity. it is so because, the gravitational intensity at a point is the gravitational force per unit mass at that point. Hence, the gravitational intensity at the centre `Q` will be also c.i.e., option (iii) is correct. |
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| 14980. |
A jet airplane is travelling at 200 ms^(-1) . The engine of the plane takes in 250 m^(3) of air of mass 32 kg each second . The air is used to burn 3 kg of fuel each second . The energy is used to compress the products of combustion to eject them at the rear of the plane at 500 ms^(-1) relative to the plane . Find the power of the jet plane in H.P.(1 H.P = 746 W). [Hint : Resultant thrust = forward thrust - backward thrust and power = force xx velocity ] |
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| 14981. |
According to Hooke's law of elasticity, if stress is increased, the ratio of stress to strain: |
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Answer» increases `("stress")/("strain")="constant"` (within the elastic LIMIT) |
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| 14982. |
A person of mass 65 kg gets into an elevator at the 30th floor of a building. The elevator begins to move at t = 0. Apparent weight of the person as a function of time is shown in figure. Take g = 10 m//s^(2) and match column -I and II: |
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Answer» If ELEVATOR MOVES DOWNWARD with acceleration a. and App. weight `= Mg + ma`. If elevator moves UPWARD with acceleration a. |
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| 14983. |
Column I shows a rigid body of circlar cross-section projected on a rough surface, with indicated velocity and angular velocity. Friction is sufficient for pure rolling. |
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| 14984. |
A block is kept the floor of an elevator at rest. The elevator starts descending with an acceleration of 12m//s^(2). Find the displacement of the block during the first 1s after the start. |
Answer» Solution : `mg-N=ma_(0)` `N=m(g-a_(0))` Here `a_(0)gtg`, the block will not be in contact with the FLOOR  For the block`:` `X=(1)/(2)g t^(2)=(1)/(2)xx10xx1^(2)=5m` In this duration, the elevator will fall a DISTANCE `s'=(1)/(2)a_(0)t^(2)=(1)/(2)xx12xxt^(2)=6m` |
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| 14985. |
A copper string is fixed between two rigid supports. The tension in the string is nearly zero, when it is just taut at 30^(@) C . Find the velocityof transverse wave in the string at 10^(@) C . For copper Young's modulusy = 1 . 3 xx 10^(11) N * m^(-2) , alpha = 1 . 7 xx 10^(-5_(@)) C^(-1) ,density =9 xx 10^(3) kg * m^(-3) . |
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| 14986. |
Two objects ofmasses in the ratio 1:2 collide elastically with each other. What is the ratio of their kinetic energies ? |
| Answer» SOLUTION :INSUFFICIENT DATA | |
| 14987. |
A body of mass 6 kg travelling with a velocity 10 m/s collides head - on and elastically with a body of mass 4 kg travelling at a speed 5 m/s in opposite direction. The velocity of the second body after the collision is |
| Answer» Answer :D | |
| 14988. |
What is the advantage of taking water in hot water bottles? |
| Answer» Solution :Water has a specifc HEAT higher than everyting but ammonia. Hence, a fixed mass for the same rise in temperature. Consequently, it LOSES more heat during cooling. This heat is used for fomentation. So a HOT water bottle remain EFFECTIVE for a LONG period of time. | |
| 14989. |
A block of weight 100N is pushed by a force F on a horizontal rough plane moves with an acceleration 1 m//s^(2) , when force is doubled its acceleration becomes, 10 m//s^(2) . The coefficient of friction is- (g=10ms^(2)) |
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Answer» `0.2` |
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| 14990. |
Molecules of CO at moderate temperature have energy equal to |
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Answer» `(7)/(2)k_(B)T` |
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| 14991. |
The velocity-time graph of a particle in one-dimensional motion is shown inFig.Which of the following formulae are correct for describing the motion of the particle over the time-interval t_(1) to t_(2) :(a) x(t_(2))=x(t_(1))+upsilon(t_(1))(t_(2)-t_(1))+(1//2)a(t_(2)-t_(1))^(2)(b)upsilon(t_(2))=upsilon(t_(1))+a(t_(2)-t_(1))(c ) upsilon_("average")=(x(t_(2))-x(t_(1)))//(t_(2)-t_(1))(d)a_("average")=(upsilon(t_(2))-upsilon(t_(1)))//(t_(2)-t_(1))(e ) x(t_(2))=x(t_(1))+upsilon_("average")(t_(2)-t_(1))+(1//2)a_("average")(t_(2)-t_(1))^(2)(f)x(t_(2))-x(t_(1))=area under the upsilon-t curve bounded by the t-axis and the dotted line shown. |
| Answer» SOLUTION :(C ), (d), (F) | |
| 14992. |
A ball is projected horizontally from a tower with a velocity of 4 ms^(-1). Find the velocity of the ball after 0.7 s. Take g = 10 ms^(-2). |
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| 14993. |
In Column-I a graph and in Column-II processes are given. Match them appropriately: |
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| 14994. |
What does the law of period of Kepler's law imply? |
| Answer» Solution :This LAW implies that as the distance of the PLANET from the SUN increases, the TIME period, ALSO increases but not at the same rate. | |
| 14995. |
The word physics is derived from ……….. Meaning nature. |
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Answer» scientist |
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| 14996. |
A ball A, moving with kinetic energy E, makes a head on collision with a stationary ball with mass n times that of A. The maximum potential energy stored in the system during the collision is: |
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Answer» `NE//(n+1)` |
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| 14997. |
Two masses m_(1) and m_(2)(m_(1)gt m_(2)) are falling from the same height when same air resistance acts on them |
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Answer» `m_(1)` has more ACCELERATION |
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| 14998. |
A river is flowing from west to east at alphaspeed of 5 metre per minute. A man on the south bank of the river, capable of swimming at 10 metre per minute in still water, wants to swim across the river by the shortest path . Calculate the direction in which he should swim. |
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Answer» Solution :The situation is shown in FIG. OA=A, represents velocity of river which is `5m mt^(-1)` OB=B , represents velocity of man in still water which is `10m mt^(-1)` The man starts swimming from O and goes along OB toreach C by shortest possible path . Here ,` /_AOC= beta =90^(@)` Here, `A=5 m mt^(-1), B=10m mt^(-1) beta = 90^(@),THETA=?` From formula , `tan beta = (B sin theta )/(A+B cos theta)` putting values, we get ,` tan90^(@)=(10 sin theta)/(5+10 cos theta)=oo (infinite)` Hence, `5+10 cos theta =0 or 10 cos theta = -5 ` (or) `cos theta =-0.5= cos 120^(@) i.e., theta = 120^(@)` It makes `/_COB=30^(@)` The man should swim along OB which is `30^(@) ` west of north . 
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| 14999. |
A piece of wax weight 18.80 g wt in air. A piece of metal is found to weight 17.03 g wt in water. It is tied to the wax and both together weight 15.23 g et, in water . Find the sp. Gravity of wax. |
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Answer» Solution :WEIGHT of wax in air = `18.80 g wt`. Weight of metal in water =`17.03 g wt`. Weight of (metal+wax) in water `= 15.23 g wt` Weight of wax in water = `15.23 - 17.03=-1.80 g wt`. Loss of weight of wax in water =`18.80 -(-1.80) = 20.60 g` wt. SP. GRAVITY of wax. =`("weight of wax in air")/("loss weight in water") = (18.80)/(20.6) = 0.91`. |
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| 15000. |
A rope passess over a pulley which is sufficiently high . Two monkeys of equel mass climb the rope from opposite ends , one of them climbing quickly than the other , relative to the rope . Assuming that the pulley and the rope are weightless and the rope is in extensible , which of monkeys will reach the top first? |
| Answer» SOLUTION :Both will reach the TOP at the same TIME because the length of the two SEGMENTS of the rope on the two sides of the pulley are always equal. | |