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.
| 38601. |
In the question number 38, the speed of the particle at this time is |
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Answer» 16 m `s^(-1)` Velocity, `VECV = (dvecr)/(dt) = (d)/(dt) (5T +1.5t^(2)) hati + 1t^(2)hatj = (5+3t)hati +2thatj` At` = 6s, vecv = 23 hati + 12 hatj` The speed of the particle is `|vecv| = SQRT((23)^(2) + (12)^(2)) = 26` m `s^(-1)` |
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| 38602. |
Assume a smooth hole drilled along the diameter of the earth. If a stone is dropped at one end it reaches the other end of the hole after a Time T_(0). Now instead of dropping the stone, you throw it into the hole with an initial velocity u. How big should u be, so that the stone appears at the other end of the hole after a time(T_(0))/(2). Express your answer in terms of acceleration due to gravity on the surface of the earth (g) and the radius of the earth (R ). |
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Answer» |
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| 38603. |
In cold countries, water pipes are laid much below the ground. Why? |
| Answer» Solution :If the pipes are LAID over ground or just below ground the pipes break when the ATMOSPHERIC temperature falls from `4^(@)`C to `0^(@)C` due to EXPANSION. | |
| 38604. |
A metal sphere of radius 1 mm and mass 50 mg falls vertically in glycerine, the viscous force exerted by the glycerine on the sphere when the speed of the sphere is 1ms^(-1) is (density of glycerine is 1260kg//m^(2) coefficient of viscosity at room temperature is 8 poise) F orliny . |
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Answer» `3XX10^(-4)N` |
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| 38605. |
Select a TRUE statement from the following |
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Answer» Year and LIGHT year have the same dimensions. |
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| 38606. |
Small drops of mercury are found to assume a spherical shape. Why? |
| Answer» Solution :Small drops of mercury are found to assume a SPHERICAL shape. This is DUE to surface of the globule being in a state of tension TRIES to CONTRACT and present the smallest area of surface. For a given VOLUME, the sphere has the minimum surface area. | |
| 38607. |
A vernier calliper has 10 divisions on vernier scale coinciding with 9 main scale divisions. It is made of a material whose coefficient of linear expansion is alpha= 10^(-3).^(@)C^(-1).At 0^(@)C each main scale division = 1mm. An object has a length of 10 cm at a temperature of 0^(@)C and its material has coefficient of linear expansion equal to alpha_(1)=1xx10^(-4).^(@)C^(-1). The length of this object is measured using the said vernier calliper when room temperature is 50^(@)C. (a) Find the reading on the main scale and the vernier scale (b) The same object is measured (at 50^(@)C) using a wooden scale whose least count is 1mm. Write the measured reading using this scale assuming it to be correct at all temperature. |
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Answer» (b) `100 PM 1`mm |
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| 38608. |
A block of mass 20 kg is pushed with a horizontal force of 90N. If the coefficient of static and kinetic friction are 0.4 and 0.3, the frictional force acting on the block is (g = 10ms^(-2)) |
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Answer» 90N |
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| 38609. |
A body is projected up with a velocity equal to 3//4^(th) of the escape velocity from the surface of the earth. The height it reaches is |
| Answer» Answer :B | |
| 38610. |
A particle performing SHM with a frequency of 5 Hz and amplitude 2 cm is initially in the positive extreme position. What is the equation for its displacement ? |
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Answer» `X = 0.02cos 10pit` |
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| 38611. |
A ball of mass .m. is rotated in a vertical by a string. The difference in tension at the top and bottom would be |
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Answer» 6mg |
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| 38612. |
A circular hole of diameter 10 cmis punched in a uniform circular lamina of diameter 26 cm such that the centre of the hole is 6 cm away from the centre of the lamina. Find the position of the centre of mass of the lamina with the hole. |
| Answer» SOLUTION :1.4 CM away from the centre of the LAMINA opposite tothe hole | |
| 38613. |
If the length of a seconds pendulum is decreased by2% find the gain or loss per day. |
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Answer» PENDULUM gains 0.0201 second in 2 second . GAIN in one DAY `= 0.0201 xx24xx60 xx60//2 = 868.32` |
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| 38614. |
In an experiment, with the position of the object fixed, a student varies the position of a convex lens are for each position, the screen is adjusted to get a clear image of the object. A graph between the object distance u and the image distance v, from the lens, is plotted using the same scale for the two axes. A straight line passing through the origin and making an an angle of 45^@ while the x-axis meets the experiemental curve at P, The coordinates of P will be : |
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Answer» `(F/2,f/2)` |
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| 38615. |
A body of mass 4kg is projected vertically up with a velocity of 20 ms^(-1). If the energy spent in over coming air resistance is 200j, the maximum height attained by it is (g= 10 ms^(-2)) |
| Answer» ANSWER :D | |
| 38616. |
Forthe mountainHimalayas, betweencentreofmassandcentreof gravitywhichoneremainsbelowandwhy? |
| Answer» Solution :sincethe HIMALAYAS areveryhighmountainsthevalueof accelerationdueto gravityat thetop of themountainsis lessthanthatat THEBOTTOM,HENCETHE CENTREOF gravityof Himalayesare lowerthanthecentreof mass . | |
| 38617. |
An object has a displacement from position vector vec(r )_(1)=(2hat(i)+3hat(j))m to vec(r )_(2)=(4hat(i)+6hat(j))m under a force vec(F)=(3x^(2)hat(i)+2y hat(j))N. Find the work done by this force. |
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Answer» Solution :`W=UNDERSET(bar(r_(1)))overset(bar(r_(f)))intvec(F).dvec(r)=underset(bar(r_(1)))overset(bar(r_(2)))int(3x^(2)hat(i)+2yhat(J)).(dxhat(i)+dyhat(j)+dzhat(k))` `underset(2)overset(4)int3x^(2)dx+underset(3)overset(6)int2y dy = [X^(3)]_(2)^(4)+[y^(2)]_(3)^(6)=83j` |
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| 38618. |
The figure below shows the plot of versus P for oxygen gas at two different temperatures Read the following statements concerning the curves given below i)The dotted line corresponds to the 'ideal'gas behaviour ii)T_(1)gtT_(2) iii)The value of (PV)/(nT) at the point ,where the curves meet on the y-axis is the same for all gases which of the above statement is true? |
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Answer» I only |
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| 38619. |
The increase in kinetic energy of a body is 69% . What will be the increases in its momentum ? |
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Answer» |
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| 38620. |
A thin uniform rod of length L and radius r rotates with an angular velocity in a horizontal plane about a vertical axis passing through one of its ends. The density of the material is rho and Youngs modulus is Y. 1) Find tension in the rod at distance Y from the axis. 2) Find elongation of the rod |
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Answer» Solution :1) Consider and element dx at a DISTANCE from the axis of rotation. 2) The centripetal force on this portion is `dt= dmr OMEGA^(2) dT=(rho A dr) R omega^(2)` This force is provided by the tension in the rod `int dT= overset(L) underset(r) int rho A dr omega^(2), T rho A omega^(2)overset(L) underset(r) intrdr, T= rho A omega^(2)[ (r^(2))/(2)]_(r)^(L)` `rArr T=(rhoA omega^(2))/(2)[ L^(2)-r^(2)] T=(1)/(2) rho A omega^(2)(L^(2)-r^(2))` (3) strain `=("stress")/(Y)` If dy is the elongation in element of length dx `(dy)/(dx)=(T)/(AY) rArr dy=(Tdx)/(AY) rArr dy=((1)/(2) rho A omega^(2)(L^(2)-r^(2)) dr)/(AY)` `int dy=(rho omega^(2))/(2y) overset(L) underset(0) int (L^(2)-r^(2))dr rArr DeltaL=(rho omega^(2))/(2y)[ L^(2)r-(r^(3))/(3)]_(0)^(L)` `DeltaL=(rho omega^(2))/(2Y)[L^(2)-(L^(3))/(3)] rArr DeltaL=( rho omega^(2))/(2y)[(2L)/(3)], DeltaL-=( rho omega^(2)L^(3))/(3Y)` |
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| 38621. |
Whatare positiondependent forces |
| Answer» SOLUTION :FORCES betweentwobodieswhichdependon the POSITIONOF thebodiesare calledpositiondependentforces EG:gravitationalforces. | |
| 38622. |
Three equal masses A, B and Care pulled with a constant force F. They are connected to each other with strings. The ratio of the tension between AB and BC is |
| Answer» SOLUTION :`F-T_(1)=m_(1)a, T_(1)-T_(2)=m_(2)a, T_(2)=m_(3)a` | |
| 38623. |
How is the force acting on a simple pendulum resolved ? |
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Answer» SOLUTION :The force is resolved into two components. The force is resolved into two components. (i) The componet `mg cos theta` is balanced by the tension in the thread acting along the length towards the fixed POINT O. (II) `mg SIN theta` whichis unbalanced, acts perpendicular to the length of thread. |
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| 38624. |
Obtain an expression for a stationary wave formed by two sinusoidal waves travelling along the same path in opposite direction, also discuss the condition for the formation of Nodes and Anti-nodes? |
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Answer» Solution :Let us consider two harmonic progressive waves (formed by strings) that have the same amplitude and same velocity but move in opposite directions. Then the displacement of the first wave (incident wave) is `y_(1)=Asin(kx-OMEGAT)""......(1)` (waves move toward right) and the displacement of the second wave (reflected wave ) is `y_(2)=Asin(kx-omegat)""......(2)` (wave move toward left) By the principle of superposition, both will interfere with each other the net displacement is `y=y_(1)+y_(2)""......(3)` Substituting equation (1) and equation (2) in equation (3), we get `{{:(Asin(kx-omegat)),(+Asin(kx+omegat)):}""......(4)` Using trigonometric identity, we rewrite equation (4) as `y(X,t)=2Acos(omegat)sin(kx)""......(5)` It represents a stationary wave or standing wave, that means that this wave does not move either FORWARD or backward, whereas progressive or travelling waves will move forward or backward. In addition, the displacement of the paricle in equation (5) can be written in more compact form, `y(x,t)=A'cos(omegat)` where, `A'=2Asin(kx)`, It is implied that the particular element of the string EXECUTES simple harmonic motion with amplitude equals to A. The maximum of this amplitude occurs at positions for which `sin(kx)=1` `implieskx=(pi)/(2),(3pi)/(2),(5pi)/(2).......` `=mpi` where m takes half integer or half integral values. The position of maximum amplitude is known as antinode. Expressing wave number in terms of wavelength, we can represent the anti-nodal positions as `x_(m)=((2m+1)/(2))(lamda)/(2)""......(6)` where m=0,1,2.... For m=0 we have maximum at, `x_(0)=(lamda)/(2)` For m=1 we have maximum at, `x_(1)=(3lamda)/(4)` For m=2 we have maximum at, `x_(2)=(5lamda)/(4)` and so on. The distance between two successive antinodes can be calculated by `x_(m)-x_(m-1)=((2m+1)/(2))(lamda)/(2)-(((2m+1)+1)/(2))(lamda)/(2)` `=(lamda)/(2)` SIMILARLY, the minimum of the amplitude A' also occurs at some points in the space, and these points can be determined by setting `sin(kx)=0` `implieskx=0,pi,2pi,3pi,.......=npi` where n takes integer or integral values. It is noted that the elements at these points do not vibrate (not move), and the points are called nodes. The `n^(th)` nodal positions is given by, `x_(n)=n(lamda)/(2)` where, n=0,1,2,..... |
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| 38625. |
What is torque? Explain the torque acting on a particle. |
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Answer» Solution :Internal force of mutual interactions among the particles of a system are equal in MAGNITUDE and opposite in direction. Therefore the resultant TORQUE due to them is zero. So we shall ignore the internal forces. Suppose `VEC(F_(1)),vec(F_(2)),vec(F_(3))....vec(F_(n))` are the external forces acting on the particles having position vectors `vec(r_(1)),vec(r_(2)),vec(r_(3))....vec(r_(n))`. The resultant torque acting on the system is the VECTOR sum of torque acting on the particles. `therefore` Resultant torue `vectau=vec(r_(1))xxvec(F_(1))+vec(r_(2))xxvec(F_(2))+......+vec(r_(n))xxvec(F_(n))` `therefore vectau=vectau_(1)+vectau_(2)+............+vectau_(n)` |
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| 38626. |
What is normal force |
| Answer» SOLUTION :Frictionalforcewhichalwaysopposethe relativemotionbetweenan OBJECTAND thesurfacewhere it isplaced. | |
| 38627. |
How can we determine whether the object is in rest or not? |
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Answer» Solution :If any object changes its position with respect to ANOTHER object, then it is said to be in motion with respect to another object OTHERWISE it is said to be in rest. For example, you are TRAVELLING in a bus moving with constant velocity and a book is there besides you. The book is stationary with respect to you. But, the book is in motion with respect to a stationary observer on road. Thus, the motion of any object is combined PROPERTY of object and observer. There is no motion without observation position. |
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| 38628. |
The motion of center of mass depends on |
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Answer» EXTERNAL FORCES acting on it |
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| 38629. |
A man starts walking from a point on the surface of earth (assumed smooth) and reaches diagonally opposite point. What is the work done by him |
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Answer» zero |
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| 38630. |
The molecular weight of oxygen and hydrogen are 32 and 2, respectively .The rms velocities of their molecules at a given temperature , will be in the ratio |
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Answer» `4:1` |
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| 38631. |
A cyclist riding with a speed of 27 kmph. As he approaches a circular turn on the road of radius 80 m, he applies breaks and reduces his speed at the constant rate of 0.50 m/s every second. The net acceleration of cyclist on the circular turn is |
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Answer» `0.5 m//s^2` |
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| 38632. |
Two simple harmonic motions with the same frequency act on a particle at right angles i.e, along x and y axis. If the two amplitudes are equal and the phase difference is (pi)/(2) the resultant motion will be : |
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Answer» a circle `Y=A sin (OMEGAT+ phi) ""....(1)` `x= A sin (omega+phi+(pi)/(2))` `x=A cos (omegat+phi) ""....(2)` On squaring and adding EQUATIONS (1) & (2) . `x^(2)+y^(2)=A^(2)` This is an equation of a circle. HENCE, RESULTING motion will be a circle motion. |
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| 38633. |
A block of mass 100 g is lying on an inclined plane of angle 30°. The frictional force on thisblock …….N. |
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Answer» `4.9xx10^(2)` Thecomponentof thisweightparallelto thesurfaceof aslopeW sin `theta` GIVESTHE frictionforce F = mgsin `theta` ![]() `=mgsin 30^(@)=0.1 xx 9.8xx (1)/(2) = 0.49 N` `f= 4.9 xx 10^(-1) N` |
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| 38634. |
The efficiency of a heat engine if the temperature of the source is 100^@C and sink is 27^@Cis nearly |
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Answer» 0.1 |
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| 38635. |
Assertion: In the siphon shown in figure, pressure at P is equal to atmospheric pressure Reason: Pressure at Q is atmospheric pressure and points P and Q are at same levels. |
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Answer» If both Assertion and REASON are true and the Reason is correct explanation of the Assertion. |
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| 38636. |
Is it possible to change the temperature of agas without supplying heat |
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Answer» Possible in ISOTHERMAL process |
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| 38637. |
A ray of light is incident on a rectangular plate at an angle of incidence 60^@. The light ray suffers a deviation which is 25% of the angle of incidence The refractive index of the glass will be |
| Answer» Answer :C | |
| 38638. |
Pressure depends on distance as, P = alpha/beta exp (-(alpha z)/(k theta))where alpha, betaare constants, z is distance, 'k' is Boltzmann's constant and thetais temperature. The dimension of betaare |
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Answer» `M^0L^0T^0` |
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| 38639. |
Two particles are executing simple harmonic motions with equal frequency and amplitude. When the displacements of the two particles are half of their amplitudes, they cross each other in mutually opposite directions. What is the phase differecne of their vibrations? |
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| 38640. |
The resistant R=V/I where V=100+-0.5V and I=10+-0.2A. The percentage error in V is 5% and in I is 2%. Find the total percentage error is R. |
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Answer» `D=m//V =10//4.4=2.27.DeltaD=0.219xxD=0.219xx2.27=0.491` M Density `=2.27+-0.491=(2.3+-0.5)GMC^(-3)`. |
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| 38641. |
A rocket is fired vertically from the earth with an acceleration of 2g, where g is the gravitational acceleration. On an inclined plane inside the rocket, making an angle thetawith the horizontal, a point object off mass m is kept. The minimum coefficient of friction mu_min between the mass and the inclined surface such that the mass does not move is: |
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Answer» `tan 2theta ` |
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| 38642. |
Calculate the work done. (a) Along path AB(b) Along path BC (c ) Along path CA (d) For the whole cycle. The indicator diagram, between P and V, is as shown. |
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Answer» <P> Solution :Word done is positive when gas expands and NEGATIVE when gas contracts.In P-V diagram, P is represented along y-axis and V is taken along x-axis. Work done is positive if the cycle is clockwise and work done is negative if cycle is anticlockwise. (a) Work done `= P xx "CHANGE in volume" Delta V` `W_(AB) = 20 xx (3-2) = 20 xx 1 = 20 J` (b) `W_(BC) = "Pressure" xx (3-3) = "zero"`. As there is no change of volume. (C ) `W_(CA)` = It is negative because volume decreases. = -Area of CALMC= -Area of (ABC + ABML) `= -((1)/(2) xx 1 xx 20 + AB xx BM)` `= -(10 + 1 xx 20) = -30 J` (d) For whole cycle, work done = -area of `Delta ABC = -(1 xx 20)/(2)` `:.` work done for whole cycle = -10 J Note : For whole cycle, we may consider `W_(AB) + W_(BC) + W_(CA)` also. |
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| 38643. |
Which of the following sets have different dimensions? |
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Answer» PRESSURE, YOUNG's modulus, stress Electric flux=electric field `xx` area `therefore` Dipole moment, electric flux and electric field have DIFFERENT dimensions. |
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| 38644. |
In the above question, the resultant upward force exerted by the sides of the glass on the water is |
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Answer» 100 N `F_(2)=(P_(a))A_(2)=(1.01xx10^(5))xx(3XX10^(-3)) =303 N` (downwards) Force exerted by bottom on the water, `F_(3)=-F_(1)=102 N` (downwards) Let F be the force exerted by side walls on THWE water (upwards) there EQUILIBRIUM of water : net upward force = net downward force or `F+F_(3)=F_(2)+W` `:.F=F_(2)+W-f_(3)=303+10-102=211 N` (downwards) |
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| 38645. |
How does the viscosity of gases depend on temperature? |
| Answer» SOLUTION :The viscosity of gases increases with the INCREASE of TEMPERATURE. | |
| 38646. |
Three equal mass m are placed at the three vertices of an equilateral triangle of side a. The gravitational force exerted by this system on another particle of mass m placed at the mid-point of a side |
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Answer» `(GM^(2))/(3a^(2))` |
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| 38647. |
A bus is moving towards a hugs wall with a velocity 5 .ms^(-1) . The driver sounds a horn of frequency 200 Hz.How many beats will a passenger of the bus hear? (Speed of sound in air = 342 ms^(-1) ) |
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Answer» |
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| 38648. |
Two forces are in the ratio of 5:2. The maximum and minimum of their resultants are in the ratio is |
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Answer» `5:2` |
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| 38649. |
A glass vessel of volume V_(o) is competely filled with a liquid and its temperature is raised by DeltaT. What volume of the liquid will overflow ? Cefficient of linear expanison of glass = alpha_(g) and coefficient of volume expension of the liquid =gamma_(1). |
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Answer» SOLUTION :Volume of the liquid overflown = Increase in the volume of the liquid ""- Increase in the volume of the container `=[V_(o)(1+gamma_(L)DELTAT)-V_(o)l]=[V_(o)(1+gamma_(g)DeltaT)-V_(o)]` `=V_(o)DeltaT(gamma_(l)-gamma_(g))=V_(o)DeltaVT(gamma_(l)-3alpha_(g))""(becausegamma_(g)=3alpha_(g))` |
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| 38650. |
Water coming out of the mouth of a tap and falling vertically in streamline flow forms a tapering column, i.e., the area of cross-section of the liquid column decreases as it moves down. Which of the following is the most accurate explanation for this ? . |
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Answer» As the water down, its speed increases and hence its pressure decreases. It is then compressed by the atmosphere. |
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