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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.
| 37701. |
Two bodies of different masses are dropped simultaneously from the top of a tower. If air resistance is proportional to the mass of the body, |
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Answer» the heavier BODY reaches the ground earlier |
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| 37702. |
Two bodies of different masses are dropped simultaneously from the topof a tower. If air resistance is same on both of them, |
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Answer» the heavier BODY REACHES the GROUND earlier |
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| 37703. |
A satellite moves around the earth in a circular orbit with speed V. If 'm' is the mass of the satellites, its total energy is |
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Answer» `-(1)/(2)MV^(2)` |
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| 37704. |
A brass disc just fits inside the hole of an iron plate. To loosen the disc from the hole easily, which of the following operations would be more convenient? (coefficient of linear expansion of brass is more than that of iron). |
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Answer» the junction should be HEATED |
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| 37705. |
A car accelerates from rest at a constant rate alphafor some time, after which it decelerates at a constant rate betaand comes to rest. If the total time elapsed is t, then the maximum velocity acquired by the car is |
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Answer» `((ALPHA^2 + BETA^2)/(alpha beta)) t` |
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| 37706. |
A bullet loses 25% of its initial kinetic energy after penetrating through a distance of 2cm in a target. The further distance it travels before coming to rest is |
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Answer» 1cm |
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| 37707. |
Three stars each at mass m, moving in a circular orbit of radius or about a stationary central star of mass M. The three identical stars orbits in the same sense and are symmetrically located with respective each other (The centre of all stars lie in one plane). Considering gravitational force of all remaing bodies on every star, the time period of each of the three stars is |
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Answer» `2pisqrt((r^(3))/(G[m + (m)/(3)]))` |
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| 37708. |
Given reasons for f_(s)=mu_(s)N |
| Answer» SOLUTION :It isnot avectorequation. THERELATION`f_(s)= mu_(s)N` is NOTA vectorrelation . Thisisbecausethe normalforceN and `f_(s)` are notinthe samedirectioneventhough`f_(s)` is equal to`mu_(s)` timesthe NORMALFORCE. thisis alsotruein the caseof KINETICFRICTION . | |
| 37709. |
A block of mass m can slide freely in a slot made in a bigger of mass M as shown in fig. There is no friction anywhere in the system. The block m is connected to one end of a string whose other end is fixed at point P. system is released from rest when the string at P makes an angle theta with horizontal. Find the acceleration of m and M after release. |
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Answer» Solution :Method 1: Just after the release of SYSTEM, block M will move towards left (say with acceleration A) and block m will move down with respect ot M (say with acceleration a). As there is no motion of M is vertical direction, hence the acceleration of m in verticle direction will be same as that of M, i.e. A, because horizontal both move together. Finding constraint Method 1: Let just after release, during a small time dt, the displacement of m(w.r.t. M) is dy downwards and that of M is dx horizontally asshown in fig. Decreases in the length of segment I`=underset(bar)d xcos theta` Increases in the length of segment II' `underset(bar)d y` As the total length of the string is constant , so `underset(bar)d = underset(bar)d x cos theta` ..(i)  Divinding both sides by dt, we get `(dy)/(dt)=(dx)/(dt) cos theta implies v = V cos theta` ..(ii) differenting w.r.t. `t=(dv)/(dt)=V(- sin theta)(d theta)/(dt)+(dV)/(dt) cos theta` But initially `V=0`, as `a=A cos theta` ...(iii) (`:. (dv)/(dt) = a , (dV)/(dt)=A`) Method 2: we can obtain (i) in the following way.  The displacment of M towards left is dx, then its component along AP is `dx cos theta`should be decreased in the length of segment I and this should be equal to increases in the length of segment II which is dy. Hence, `dy=dx cos theta`. `rArr a = A cos theta` Method 3: Total length of string : `l=sqrt(h^(2)+y^(2))+x` differentiating w.r.t. `t=(dl)/(dt) =(d(h^(2)+x^(2))^(1//2))/(dt) +(dx)/(dt)` `rArr 0=(2y(dy)/(dt))/(2sqrt(h^(2)+y^(2)))+(dx)/(dt)rArr0= cos theta (-V)+r` (`because(dx)/(dt)=v, (dy)/(dt)=-V`(as y decreases with time) and `cos theta=(y)/(sqrt(h^(2)+y^(2)))` `implies v=V cos theta` which is same as (ii) Now finding acceleration of blocks We can use any of the following method to find the value of A and a .  Method 1: From FBD of M: `T cos theta-N = MA` ...(IV) From FBD of m: `N=mA` ..(v) `MG-t=ma` ..(vi) Combining (iv) and (v) we get `T cos theta=(m+M)A`...(vii) Putting `a=A cos theta` in (iv) we get `mg-T=mA cos theta`..(viii) Put the value of T form (viii) into (vii) we get `(mg-mA cos theta)cos theta=(m+M)A` `implies A=(mg cos theta)/(M+m(1+cos^(2)theta))` and `a=A cos theta=(mg cos^(2)theta)/(M+m(1+cos^(2) theta))` The net acceleration of m is `sqrt(A^(2)+a^(2))=A sqrt(1+cos^(2) theta)=(mg cos thetasqrt(1+cos^(2) theta))/(M+m(1+cos^(2) theta))` Method 2: Analyzing the block with w.r.t. wedge  W.r.t M, m has only vertcal acceleration which is a downwards. Here we have to apply pseudo force because M is a non inertial frame. FBD of m w.r.t. M: `N=mA, mg-T=ma` These equation are same as (v) and (vi). Now solve to get answer. Method 3: Analyzing the block and wedge together Analyzing the system of `(M+m)`, i.e. , block and wedge together in horizontal direction, in horizontal direction both masses have same acceleration which is A.  `T cos theta=(M+m)A` ...(ix) Analyzing the motion of blcok in vertical direction `mg-T=ma` ..(x)  SOLVING (ix) and (x) , we get `A=(mg cos theta)/(M+m(1+cos^(2) theta))` and `a=(mg cos^(2)theta)/(M+m(1+cos^(2) theta))` |
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| 37710. |
The Young's modulus of steel is twice that of brass. Two wires of the same length and of the same area of cross section, one of steel and another of brass are suspended from the same roof. If we want the lower ends of the wires to be at the same level, then the weights added to the steel and brass wires must be in the ratio of |
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Answer» `1:1` |
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| 37711. |
Mass of the earth is 81 times that of the moon. If the distance between their centres is d, then the point on the line joining their centres at which the gravitational field due to them is zero is |
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Answer» `(d)/(10)` from the CENTRE of MOON |
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| 37712. |
"The rotational vectors are represented by straight lines lying alongaxis of rotation". Why ? |
| Answer» SOLUTION :Because they are AXIAL VECTORS, They RESULT from the vector product of TWO vectors. | |
| 37713. |
The efficiency of Carnot heat engine is 0.5 when the temperature of the source is T_(1) and that of sink is T_(2). The efficiency of another Carnot's heat engine is also 0.5 . The temperature of source and sink of the second engine are respectively |
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Answer» `2T_(1),2T_(2)` |
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| 37714. |
A bullet strikes the wooden block and gets embedded into it. What type of collision is it? |
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Answer» PERFECTLY ELASTIC |
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| 37715. |
The mass and diameter of a planet are twice that of the earth. What will be the time period of oscillation of a pendulum on this planet, if it is seconds pendulum on earth. |
| Answer» Answer :D | |
| 37716. |
The velocity of Boat in still water is 3m/s. Velocity river is 4m/s. Walking velocity of man on Bank is 1 m/s {:("Column-I","Column-II") ,("A) Shortest time taken to cross river ","P) " 40 sqrt(7) "SI units"),("B) Time taken to cross river in shortest path","Q) " (40)/(sqrt(7)) "SI units"),("C) Shortest time taken to reach directly opposite point","R) 40 SI units"),("D) Minimum drift","S) 160 SI units"),(,"T)" (160)/(sqrt(7))"SI units"):} |
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Answer» |
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| 37717. |
A particle initially at rest starts moving from point A on the surface of a fixed smooth hemisphere of radius r as shwon. The particle looses its contact with hemisphere at point B. C is centre of the hemisphere. The equation relatin alpha and beta is |
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Answer» `2sinalpha=2cosbeta` |
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| 37718. |
Figure shown a cylindrical tube of radius r and lengh l fitted with a cork. The friction coefficient between the cork and the tube is mu. The tube contains an ideal gas at temperature T, and atmospheric pressure P_(0) Thetube is slowly heated , the cork pipe out when temperature is boubled. What is normal force per unit length exerted by the periphery of tube? Assume uniform temperature throughout gas any instant. |
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Answer» <P> Solution :SINCE volume of the gas is constant,`(P_(i))/(T_i)=(P_f)/(T_(f)` `P_(f)=P_(i)((T_(F))/T_(i))=2P_(i)=2P_(0)` The Forces actihg on the CORK are shown in the FIGURE in equilibrium. `P_(0)xxA+muN=2P_(0)A` `N=(P_(0)A)/mu` `N` is the total normal force exerted by the tube on the cork, hence contact force per unil length is `(dN)/(dl)=(N)/(2pir)`=`(P_(0)A)/(2pimur)` |
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| 37719. |
A solid sphere of radius ‘r’ is kept on a concave mirror of radius of curvature R. The arrangement is kept on a horizontal table. If the sphere is displaced from its equilibrium position and left, then it executes SHM. The period of oscillation will be (sphere does't slip) |
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Answer» `2PI SQRT([(R - r)(1.4)G])` |
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| 37720. |
A source of sound moves towards an observer |
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Answer» The FREQUENCY of the source is increased |
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| 37721. |
A saturn year is 29.5 times the earth year. How far is the saturn from the sun if the earth is 1.50 × 10^(8) km away from the sun ? |
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Answer» Solution :`implies` TIME period of earth = 1 yr Time period of planet Saturn = 29.5 yr ORBITAL radius of earth `R_E = 1.5 XX 10^(8)`km Orbital radius of planet Saturn `= R_S` According to Kepler.s THIRD law, `T^(2) prop r^3` `:. ((T_E)/(T_s))^2 = (R_E/R_S)^3` `:. (1/(29.5))^2 =((1.5xx10^8)/(R_s))^3` `:. R_s^3 = (29.5)^2 (1.5 xx10^8)^3` `= 2.947 xx 10^(27) km` `:. R_s= 1.43 xx10^(9) km` `= 1.43 xx10^(12) m` |
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| 37722. |
If look inside a real piano , you'll see that the assumption made in part (b) of Illustration 7.31 is only partially true . The strings are not likely to have the length of the A string is only 64 % of the length of the C string . What is theratio of their tensions ? |
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Answer» Solution :The ratio of frequencies : `( f_(1) A)/( f_(1) C) = (L_(C ))/(L_(A)) sqrt (( T_(A))/( T_(C)))` `(T_(A))/(T_( C)) = ((L_(A))/(L_(C )))^(2) ((f_(1) A)/( f_(1) C))^(2)` `(T_(A))/( T_( C )) = ( 0.64)^(2) (( 440)/( 262))^(2) = 1.16` NOTICE that this result REPRESENTS only a `16 %` increase in tension , compares with the `18.2%` increase in PART (b) of Illustration 7.31. |
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| 37723. |
The cylindrical tube of a spray pump has a cross section of 8.0 cm^2 one end of which has 40 fine holes each of diameter 1.0 mm. If the liquid flow inside the tube is 1.5 m min^-1. What is the speed of ejection of the liquid through the holes? |
| Answer» SOLUTION :`0.64 ms^-1` | |
| 37724. |
As shown in the figure, a planet revolves round the sun in Elliptical orbit with sun at the focus. The shaded areas can be assumed to be equal. If t_(1) and t_(2) represent the time taken for the planet to move from A to B and C to D respectively, then |
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Answer» `t_(1) LT t_(2)` |
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| 37725. |
The maximum number of components a vector can be split are ? |
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Answer» 2 |
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| 37726. |
At what temperature will the average velocity of oxygen molecules be sufficient so as to escape from the earth ? Escape velocity of earth is 11.0 kms^(-1) and mass of one molecule of oxygen is 5.34xx10^(-26) kg. Boltzmann constant =1.38xx10^(-23) " J molecule"^(-1) K^(-1) . |
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Answer» SOLUTION :For the MOLECULE to just escape from the earth, AVERAGE KINETIC energy of the molecule at temperature T = Escape energy of the molecule. (or) `3/2k_BT=1/2mv_e^2` `T = (mv_e^2)/(3k_B)` `=(5.34xx10^(-26)xx(11.0xx10^3)^2)/(3xx1.38xx10^(-23))` `= 1.56xx10^5 K`. |
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| 37727. |
A sphere of aluminium of 0.047 kg placed for sufficient time in a vessel containing boiling water, so that sphere is at 100" "^(@)C. It is then immediately transferred to 0.14 kg copper calorimeter containing 0.25 kg of water at 20" "^(@)C. The temperature of water rises and attains a steady state at 23" "^(@)C. Calculate the specific heat capacity of aluminium. |
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Answer» Solution :Mass of sphere `m_(1)=0.047` KG Initial temperature of sphere `T_(1)=100^(@)C` Final temperature of sphere `T_(2)=23^(@)C` Change in temperature `DeltaT_(1)=T_(1)-T_(2)` `=(100-23)" "^(@)C` `=77^(@)C` Sphere specific heat of sphere `=S_(1)` Heat lost by aluminum sphere, `Q_(1)=m_(1)S_(1)DeltaT_(1)` `=0.047xxS_(1)xx77` `:.Q_(1)=3.619S_(1)`. . .(1) Now, mass of water `m_(2)=0.25` kg Mass of calorimeter `m_(3)=0.14` kg Specific heat of water, `S_(2)=4.18xx10^(3)" Jkg"^(-1)K^(-1)` Specific heat of copper calorimeter, `S_(3)=0.386xx10^(3)" Jkg"^(-1)K^(-1)` Heat gained by water and calorimeter, `Q_(2)=m_(2)S_(2)DeltaT_(2)+m_(3)S_(3)DeltaT_(2)` `=0.25xx4.18xx10^(3)xx3+0.14xx0.386xx10^(3)xx3` `=3.135xx10^(3)+0.16212xx10^(3)` `:.Q_(2)=3.29712xx10^(3)`. . .(2) For STEADY state, {Heat lost by sphere} = {Heat gained by water} + {Heat gained by calorimeter} `Q_(1)=Q_(2)` `3.619S_(1)=3.29712xx10^(3)` [from equation (1) and (2)] `:.S_(1)=(3.29712xx10^(3))/(3.619)=0.911xx10^(3)` `:.S_(1)=0.911" kJ kg"^(-1)=911" Jkg"^(-1)K^(-1)` (specific heat of aluminium) |
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| 37728. |
(1)/(12) of the mass of carbon 12 atom is...... |
| Answer» Solution :mass of hydrogen | |
| 37729. |
A simple pendulum excuting S.H.M in a straight line has zero velocity at 'a' and 'b' whose distances from 'o' in the same line OAB are 'a' and 'b'. If the velocity half-way between them is 'v' then its time period is |
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Answer» `(PI(B+a))/(V)` |
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| 37730. |
An aircraft executes a horizontal loop of radius 1.00 km with a steady speed of 900km/h. Compare its centripetal acceleration with the acceleration due to gravity . |
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Answer» |
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| 37731. |
A siren of frequency n approaches a stationary observer and then receedes from the observer. If the velocity of source (V) lt ltthe velocity of sound (C ), the apparent change in frequency is |
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Answer» `2nV//C` `n..=n((C )/(C+V))` `Deltan..=n.-n..=NC[(1)/(C-V)-(1)/(C+V)]` `=nC[(2V)/(C^(2)-V^(2))]` `V^(2)`can be NEGLECTED as `V lt lt C :. Deltan.=(2nV)/(C )` |
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| 37732. |
A bullet of mass 10gm hits a fixed target and penetrates through 8cm in it before coming to rest. If the average resistance offered by the target is 100N, the velocity with which the bullet hits the target is, |
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Answer» 20m/s |
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| 37733. |
A body of mass (4m) is lying in x-y plane at rest. It suddenly explodes into three pieces. Two pieces, each of mass (m) move perpendicular to each other with equal speeds (v). The total kinetic energy generated due to explosion is |
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Answer» `mv^2` |
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| 37734. |
We can cut vegetables easily with a sharp knife as compared to a blunt knife. Why ? |
| Answer» SOLUTION :The area of a sharp edge is much Less than the area of blunt edge. For the same total force, defective force PER UNIT area is more for the sharp edge than the blunt edge. Has a sharp knife CUTS EASILY than a blunt knife. | |
| 37735. |
Two particles of masses 4 kg and 6 kg are at rest separated by 20 m. If they move towards each other under mutual force of attraction, the position of the point where they meet is |
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Answer» `12M` from `4KG` body |
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| 37736. |
Below are four equations expressing relationships between physical qualities. .g. gravitational field strengths, r radius, rho a density, G the universal constant of gravitation, m a mass and T a time period. Which of the are true. equations are true. (a)gT^2=4pir^2 , (b)4/3pirhoGr^2=g , (c)gr^2=Gm , (d)g=4/3 pirhoGr |
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Answer» only a&B are true |
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| 37737. |
The bob of a simple pendulum of mass 200g is oscillating with a time 2sec. If the bob replaced by another bob of mass 450g but of same radius what will be the new time period of oscillation ? |
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Answer» 1sec |
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| 37738. |
A boy of mass 30 kg is running with a velocity of 3 ms^(-1) on ground just tangenti - ally to a merry - go - round which is at rest. It has radius R = 2m, a mass of 120 kg and its radius of gyration is 1m. If the boy suddenly jumpes on to the merry - go - round, the angular velocity acquired by the system is |
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Answer» `1 RAD s^(-1)` |
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| 37739. |
The first law of thermodynamics is a law of conversion of |
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Answer» ENEGY |
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| 37740. |
Whatis meantby thermalequilibrium ? |
| Answer» Solution :TWO SYSTEMS are said to be in thermal equilibrium with each other if they are at the same TEMPERATURE, which will not change with TIME. | |
| 37741. |
What are difficulties arised by using water in a barometer instead of mercury ? |
| Answer» SOLUTION :The DENSITY of water is low , hence if water is used in BAROMETER the LENGTH of tube used in barometer should keep very long (about 11 meter ) .So it is not convenient . | |
| 37742. |
Mercury has an angleof contact equal to 140^(@) with soda lime glass . A narrow tube of radius 1.00 mm made of this glass is dipped in a trough containing mercury .By what amount does the mercury dip down in the tube relative to the liquid surface outside ?Surface tension of mercury at the temperature of the experiment is 0.465Nm^(-1) . Density of mercury =13.6xx10^(3)kgm^(-3) |
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Answer» Solution :Here , angle of CONTACT `theta=140^(@)` Radius of NARROW tube `r=1mm=10^(-3)m` Surface TENSION `T=0.465Nm^(-1)` DENSITY of mercury `rho=13.6xx10^(3)kgm^(-3)` Height of liquid in narrow tube of radius r, `h=(2Tcostheta)/(rrhog)` `=(2xx0.465xxcos140^(@))/(10^(-3)xx13.6xx10^(3)xx9.8)` `=0.00534498` `=-5.34xx10^(-3)m` Negative sign shows that the mercury level is depressed in the tube. |
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| 37743. |
A bus starts from rest with a constant acceleration . At the same time a car travelling witha constantvelocity 50 m/s over-takes and passes the bus. How fast is the bus travelling when they are side by side ? |
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Answer» 10 m/s |
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| 37744. |
Assuming the earth to be a sphere of uniform mass density, how much would a body weigh half way down to the centre of the earth if it weighed 250 N on the surface ? |
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Answer» Solution :`IMPLIES` Suppose, G and `g_d`are the acceleration DUE to gravity at the surface and at the depth d of earth respectively and `W and W_d` are the weights of body at the surface and at the depth of earth respectively. `g(d) = g(1-d/R)` `:. g(d) = g (1-(R/2)/R)""[ :. d = R/2]` `= g (1-1/2)` `=g/2` `:.`Weight of body of mass m at depth d W(d) = mg (d) `=(mg)/2 ""[ :. g(d) =g/2]` `W(d) = W/2 ""[ :. mg = W]` `:. W(d) = (250)/2` `:. W(d) = 125 N` |
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| 37745. |
A copper ball cools from 62^(@)C to 50^(@)C in 10 minutes and to 42^(@)C in the next 10 minutes. Calculate its temperature at the end of another 10 minutes. |
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Answer» SOLUTION :`(theta_(1)-theta_(2))/t_(1)=K[(theta_(1)+theta_(2))/2-theta_(0)]` `(62-50)/10=k[(62-50)/2-theta_(0)]" "...(1)` `(theta_(2)-theta_(3))/t_(2)=k[(theta_(2)+theta_(3))/t_(2)-theta_(0)]` `(50-42)/10=k[(50+42)/(2)-theta_(0)]""...(2)` `((1))/((2))" gives "theta_(0)=26^(@)C` `(theta_(3)-theta_(4))/t_(3)=k[(theta_(3)+theta_(4))/2-theta_(0)]` `(42-theta_(4))/10=k[(42-theta_(4))/2-26]rArrtheta_(4)=36.7^(@)C` |
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| 37746. |
When 1 gm of water changes from liquid to vapour phase at constant pressures of 1 atmosphere, the volume increases from 1 cc to 1671 cc. The heat of vaporisation at this pressure is 540 cal/gm. Increase in internal in internal energy of water is (1 atmosphere = 1.01xx10^(6) "dyne/cm"^(2)) |
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Answer» 4200 J |
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| 37747. |
Calculate the black body temperature of the sun from th e following data :- Stefan's constant = 5.7xx10^(-8) Wm^(-2)K^(-4) . Solarconstant = 1500 W//M^(2). Radius of the sun = 7xx10^(8)m . Distance between sun and earth = 1.5xx10^(11) m. |
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Answer» SOLUTION :`sigma =5.7xx10^(-8)Wm^(-2)K^(4), S = W//m^(2) , R =7xx10^(8) m, r =1.5xx10^(11) m` `T = [(r/R)^(2)S_(0)/sigma]^(1//4) = [((1.5xx10^(11))/(7xx10^(8)))^(2)xx1500/(5.7xx10^(-8))]^(1/4)`= 5895 K |
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| 37748. |
A rain drop of mass m, starts falling from rest and it collects water vapour and grows. If it gains lambda kg/s, find its velocity at any instant. |
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Answer» Solution :`(d)/(dt) (MV) = mg , (dm)/(dt) = lambda or m= lambda t + K` where K is a constant when t=0 , `m=m_0 :. m=m_0 + lambda t ` so , ` (d)/(dt) {(m_0 + lambda t)v} = (m_0 + lambda t ) g` Integrating , `(m_0 + lambda t) v= int (m_0 + lambda t) g dt = (m_0 t + ( lambda t^(2))/(2) ) g + C ` where C is constant when t =0, v=0 ` :.` C =0 `:. (m_0 + lambda t) v = ( m_0 t + ( lambda t^2)/(2)) g , v=(g(m_0 t + (lambda t^2)/( 2)))/(m_0 + lambda t ) =(g (t+ (lambdat^(2))/(2m_0)))/( 1 + (lambda t)/( m_0))` This mean VELOCITY changes w.r.t . time |
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| 37749. |
Two wires of same length and radius are joined end to end and loaded. The Young.s modulii of the materials of the two wires are Y_(1) and Y_(2). If the combination behaves as a single wire them its Young.s modulus is |
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Answer» Solution :`e=e_(1) +e_(2)" but " e=(F(2L))/(AY_(EQ)), e_(1)=(FL)/(AY_(1)), e_(2)=(Fl)/(AY_(2))` `(F(2l))/(AY_(eq))=(Fl)/(AY_(1))+(Fl)/(AY_(2))` `(2)/(Y_(eq))=(1)/(Y_(1))+(1)/(Y_(2)) RARR Y_(eq)=(2Y_(1)Y_(2))/(Y_(1)+Y_(2))` |
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| 37750. |
Find the component of 3hat(i)+ 4hat(j) along hat(i)+ hat(j) |
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Answer» SOLUTION :Component of `VEC(A) "ALONG" vec(B)` is GIVEN by `(vec(A).vec(B))/(B)= ((3 hat(i) + 4hat(j)).(hat(i) + hat(j)))/(sqrt2)= (7)/(sqrt2)` |
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