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.
| 39101. |
Find the depth at which the value of g becomes 25% of that at the surface of the earth. (Radius of the earth = 6400 km) |
|
Answer» SOLUTION :G at a depth `g_(d) = g(1-(d)/(R))` In this problem, `g_(d) = (25)/(100)g = 0.25 g` SUBSTITUTING `g_(d) = 0.25g` and R = 6400 km we GET d = 4800 km |
|
| 39102. |
In damped oscillatory motion a block of mass 200g is suspended to a spring of force constant 90 N//m in a medium and damping constant is 40g//s. Find time taken for its mechanical energy to drop to half of its initial value. |
|
Answer» Solution :Mass ,= 200g= 0.2 kg force constant k = `90N//m` DAMPING constant `b = 40g//s =0.04 kg//s`. `sqrt(km)= sqrt(90 xx 0.2)= sqrt(18) kg//s` Here `b lt lt sqrt(km)` Let the ENERGY is dropped to HALF of its initial value after a time `t_(1//2)` intial energy `E_(0)= (1)/(2)KA^(2)` At time `t_(1//2)` , energy =`(1)/(2)E_(0)= (1)/(2)kA^(2)e^((-bt_(1//2))/(m))= (1)/(2)((1)/(2)kA^(2))` `e^((-bt_(1//2))/(m))= (1)/(2) implies t_(1//2)= LN(2) xx (m)/(b) = 0.693 xx (m)/(b)` `t_(1//2)= 0.693 xx (0.2)/(0.04)= 3.46s` |
|
| 39103. |
The dimensional formula for angular momentum is |
| Answer» SOLUTION :ANGULAR MOMENTUM is CONSTANT | |
| 39104. |
Suppose you keep your two hands on two bodies simultaneously and feel them equally cold or hot. What is your inference? |
| Answer» SOLUTION :They are at the same TEMPERATURE as your BODY (THERMAL EQUILIBRIUM) | |
| 39105. |
The coefficients of viscosity of two liquids are in the ratio 2:3. What is the ratio of the volumes of liquid collected in the same time using the same vessel completely filled with liquids of densities in the ratio 4:5? (The capillary tubes used have the same length and same bore diameter) |
|
Answer» `8:15` |
|
| 39106. |
ABCDEF is a regular hexagon with point O as centre. Find the value of vec(AB) + vec(AC) + vec(AD) + vec(AE) + vec(AF) |
|
Answer» `6 vec(OA)` |
|
| 39107. |
A bullet moving with a uniform velocity V stops suddenly after hitting a target and the whole mass melts. If the mass of the bullet is m, specific heat is S, initial temperature is 25^(0)C, melting point is 475^(0)C and latent heat is L, find the velocity of the bullet in terms of L and S in SI units. |
|
Answer» |
|
| 39108. |
Three particles of masses 1 kg, 2 kg, 3 kg are acted upon by forces (veci +2vecj),(2vecj +3veck) and (veci - veck) newton respectively. The magnitude of acceleration of centre of mass in m/s^2 is |
|
Answer» `SQRT(3)` |
|
| 39109. |
A cylindrical block of length 0.4 m and area of cross-section 0.04m^(2) is placed coaxially on a thin metal, disc of mass 0.4 kg and same cross-section. The upper face of the cylinder is maintained at a constant temperature of 400K and the initial temperature of the disc is 300K. If the thermal conductivity of the material of the cylinder is 10W/mK, the specific heat of the material of the disc is 600 J/kg-K, and the time taken for the temperature of the disc to increase to 350K is x min 48sec, then what is the value of x? Assume for purpose of calculation the thermal conductivity of the disc to be very high and the system to bee thermally insulated except for the upper face of the cylinder. |
|
Answer» |
|
| 39110. |
A particle of mass m_(1) (A) elastically collides with another stationary particle (2) of mass m_(2). Then : |
|
Answer» `(m_(1))/(m_(2)) = (1)/(2)` and the particles fly apart in the opposite direction with equal velocities. |
|
| 39111. |
Show that simple harmonic motion may be regarded as the projection of uniform circular motion along a diameter of the circle. |
|
Answer» Solution :Tie a ball to the end of a STRING and make it move in a HORIZONTAL PLANE about a fixed point with a constant angular speed. The ball then perform cirular motion in the horizontal phane. Observe the motion of ball in a plane by lighting torch in a dark room. The ball will appear to execute to and fro motion along a horizontal line with the point of rotation as the mid-point. Also observe the shadow of the ball on a wall which is perpendicular to the plane of the circle. Here the motion of the ball on a DIAMETER of the circle normal to the direction of viewing. This is shown in below figure.  Suppose position of ball moving in circular motion are ABCDEFGA. Observing this plane perpendicularly, the shadow of ball appear on wall will be OPQPORSRO respectively. The motion of this shadow seems AROUND point O (up and down) which is simple harmonic motion. |
|
| 39112. |
A particle moves along the X-axis according to to the law S=a sin^(2)(omegat-pi//4) The velocity of the particle (mu), as a fiction of the co-ordinate 'x' is |
|
Answer» `u^(2)=4w^(2)x(a-x)` |
|
| 39113. |
The maximum and minimum tensions in a string of length 0.1m with a stone 5g tied to it and whirled in a vertical circle are in the ratio 13:1. The velocity of the body at highest point is: |
|
Answer» `0.7 ms^(-1)` |
|
| 39114. |
Two plano concave lens of glass of refractive index 1.5 have radii of curvature of 20 and 30 cm. They are placed in contact with curved surfaces towards each other and the space between them is filled with a liquid of refractive index (4//3). Find the focal length length of the system. |
|
Answer» Solution :As shown in the FIGURE the system is equivalent to combination of three THIN lenses in CONTACT i.e., `1/F=1/f_1+1/f_2+1/f_3` But by LENS maker.s formula `1/f_1=[3/2-1][1/(-infty)-1/20]=-1/40` `1/f_2=[4/3-1][1/20+1/30]=5/180` `1/(f_3)=[3/2-1][1/(-30)-1/(-infty)]=-1/60 so, 1/f=-1/40+5/180-1/60` `i.e., 1/F=(-90+10-6)/360 or F=-72cm` i.e., the system will behave as a divergent lens of focal LENGTH 72cm. 
|
|
| 39115. |
There is always an excess of pressure inside drops and bubbles. Two soap bubbles A and B are blown at the end of a tube, as shown below. Choose the correct answre: When the block C is removed, ......... |
|
Answer» the SIZE of A INCREASES and that of B decreases. |
|
| 39116. |
Give two uses of Pascal's law |
|
Answer» SOLUTION :The devices WORKING on Pascal.s law : (1) Hydraulic LIFT (2) Hydraulic brakes |
|
| 39117. |
four pairs of initial and final positions of a body along an x-axis are given. which pair gives a positive displacement of the body? |
|
Answer» `-10m,+15M` |
|
| 39118. |
You have learnt that a travelling wave in one dimension is represented by a function y= f(x, t) where x and t must appear in the combination x-v t or x + v t, i.e., y= f (x +- v t). Is the converse true? Examine if the following functions for y can possibly represent a travelling wave: (a) (x-vt)^(2) (b) log [(x + vt)//x_(0)] (c ) 1//(x + vt) |
| Answer» SOLUTION :The converse is not TRUE. An obvious requirement for an ACCEPTABLE function for a travelling wave is that it should be finite everywhere and at all times. Only function (c ) SATISFIES this condition, the REMAINING functions cannot possibly represent a travelling wave. | |
| 39119. |
Determine the volume of water flowing per second through a hole of cross sectional area 0.5 cm^2 situated at a depth of 2.5 m from the surface in a tank of water. |
| Answer» SOLUTION :`0.35L.s^-1` | |
| 39120. |
The above graph represents temperature versus heat for water at 1 atm. Pressure. The slopes of AB and CD are different. Why ? |
| Answer» SOLUTION :Because the specific HEAT CAPACITY of water is GREATER than that of ice. | |
| 39121. |
The outer side of a circular track of radius 200 m is raised to make an angle of 15^@ with the horizontal Using the data provided in this case,determine the maximum permissible speed to avoid skidding. (Given mu_s = 0.25). |
|
Answer» Solution :`R = 200 m, theta = 15^@, mu_s = 0.25` `v_max = SQRT Rg(TAN theta + mu_s)/1- mu_s tan theta = sqrt 200 XX 9.8(tan 15^@ + 0.25/1-0.25 xx tan 15^@= 32.98 MS^(-1)` |
|
| 39122. |
The outer side of a circular track of radius 200 m is raised to make an angle of 15^@ with the horizontal. Name the process by which the outer side of a curved track is raised a littile above the innerside. |
| Answer» SOLUTION :BANKING of ROAD | |
| 39123. |
The outer side of a circular track of radius 200 m is raised to make an angle of 15^@ with the horizontal. Which force provides the necessary centripetal force for a car taking the circular track ? |
| Answer» SOLUTION :In this case, the COMPONENTS of NORMAL reaction and the frictional force TOWARDS the centre provides necessary CENTRIPETAL force. | |
| 39124. |
The mean kinetic energy of 1 mole of gas per degree of freedom is |
|
Answer» `1/2k_B T` |
|
| 39125. |
Can the value of the coefficient of friction be greater than 1? |
| Answer» Solution :Value of the COEFFICIENT of friction (`mu`) is usually less than 1. But is some SPECIAL cases its value can be equal to or even greater than 1. The value of `mu` between TWO metal surfaces , cleaned scientifically and KEPT in vacuum may rise up to 10 approximately. Under these conditions `mu ~~1.6` between two copper plates. | |
| 39126. |
If an object reaches a maximum vertical height of 23.0 m when thrown vertically upward on earth how high would it travel on the moon where the acceleration due to gravity is about one sixth that on the earth ? Assume that initial velocity is the same. |
|
Answer» Solution :`H alpha 1/G h_(1)/h_(2)=g_(2)/g_1=1/6` `RARR h_2=6h_(1)` `6 XX 23=138m` |
|
| 39127. |
A longitudinal wave is described by the equation y = y_(0) sin 2pi (ft - x// lambda) . The maximum particle velocity is equal to four times the wave velocity if ……….. . |
|
Answer» `lambda = pi y_(0)//4` `v_("max") = 2pi fy_0` `v_("max") = 2pi f y_(0) = 4 f lambda` `implies lambda = (pi y_(0))/(2)` |
|
| 39128. |
A horizontal tube of length l closed at both ends contains an ideal gas of molecular weight M. The tube is rotated at a constant angular velocity m about a vertical axis passing through an end, Assuming the temperature to be uniform and constant, pressure P at the free end is___ (pressure at fixed end is P_0 ) |
|
Answer» `P = P_(0) e^( M omega^(2) t^(2))/( 2 RT)` |
|
| 39129. |
Sliding of the object occurs while |
|
Answer» `V_("TRANS")= V_("ROT")` |
|
| 39130. |
The velocity of a body moving vertically up is 49ms^(-1) at half the maximum height. The height to which it could further rise is |
|
Answer» 245m |
|
| 39131. |
If the kinetic energy of a body is fourtimes its momentum, its velocity is |
| Answer» Answer :C | |
| 39132. |
A force F is related to the position of a particle by the relation F = (10x^2) N. The work done by the force when the particle moves from x= 2 m to x = 4 m is |
|
Answer» `56/3 J` `= 10/3 [4^3 - 2^3] = 10/3 xx 56 = 560/3 J` |
|
| 39133. |
Figure given below shows the variation of potential energy function U(x), corresponding to one dimensional force field. The point on the graph representing the position of most stable equilibrium is : (##NAR_NEET_PHY_XI_P2_C06_E09_112_Q01.png" width="80%"> |
|
Answer» A |
|
| 39134. |
Two rigid bodies have same moment of inertia about their axes of symmetry. Which will have greater kinetic energy? |
|
Answer» Solution :Relation between angular MOMENTUM and kinetic energy is KE = `(L^(2))/(2I)` Even THOUGH moment of inertia is same, the BODY with larger angular momentum will have larger kinetic energy |
|
| 39135. |
A flask contains argon and chlorine in the ratio of 2:1 by mass. The temperature of the mixture is 27^(@)C. Obtain the ratio of (i) average kinetic energy per molecule, and (ii) root mean square speed v_("rms") of the molecules of the two gases.Atomic mass of argon =39.9 u, Molecular mass of chlorine = 70.9 u. |
|
Answer» Solution :The important point to REMEMBER is that the average kinetic energy (per molecule) of any (ideal) GAS (be it monatomic LIKE argon, diatomic like chlorine or polyatomic) is always equal to `(3//2) k_(B)T`. It depends only on temperature, and is independent of the nature of the gas. (i) SINCE argon and chlorine both have the same temperature in the flask, the ratio of average kinetic energy (per molecule) of the two gases is 1:1. (ii) Now `(1)/(2)v_("rms")^(2)=`average kinetic energy permolecule =`(3//2) k_(B)T` where m is the mass of a molecule of the gas. Therefore, `(V_("rms")^(2))_(Ar)/(V_("rms")^(2))_(Cl)= (m)_(Cl)/(m)_(Ar)=(M)_(Cl)/(M)_(Ar)=(70.9)/(39.9)= 1.77` where M denotes the molecular mass of the gas. (For argon, a molecule is just an atom of argon.) Taking square root of both sides, `(V_("rms"))_(Ar)/(V_("rms"))_(Cl)= 1.33` You should note that the composition of the mixture by mass is quite irrelevant to the above Calculation . any other propotion by mass of argon and chlorine would give the same answers to (i) and (ii) provided the temperature REMAINS unaltered. |
|
| 39136. |
What is elastic limit? |
| Answer» SOLUTION :The maximum value of STRESS UPTO which MATERIAL will have its elastic PROPERTY. | |
| 39137. |
Which of the following relationship between the acceleration a and the displacement x of a particle involve simple harmonic motion ? |
|
Answer» `a=0.7x` |
|
| 39138. |
Passage - III : Four identical spheres having mass M and radius R are fixed tightly within a massless ring such that the centres of all sphere lie in the plane of ring. The is kept on a rough horizontal table as shown. The string is wrapped around the ring which can roll without slipping. The other end of the string is passed over a massless frictionless pulley to a block of mass M.A force F is applied horizontally on the ring, at the same level as the centre, so that the system is in equilibrium. If the masses of the sphere were doubled keeping their dimension same, the force of friction between the ring and the horizontal surface would |
|
Answer» Be DOUBLED |
|
| 39139. |
Passage - III : Four identical spheres having mass M and radius R are fixed tightly within a massless ring such that the centres of all sphere lie in the plane of ring. The is kept on a rough horizontal table as shown. The string is wrapped around the ring which can roll without slipping. The other end of the string is passed over a massless frictionless pulley to a block of mass M.A force F is applied horizontally on the ring, at the same level as the centre, so that the system is in equilibrium. The moment of inertia of the combined ring system about the centre of ring and magnitude of F equals to |
|
Answer» `(12)/(5)MR^(2),MG` |
|
| 39140. |
Assertion :Whencamphoris burnt it vapourisesduringheating.thatissolidis converted intogas in thiscase.Thisis calledsublimation. Reason.calorimetrymeansthemeansurementof the amountof heatreleasedor absorbedby thhermodynamicsystemduringthe heatingprocess :It can beexpressed as Q_("gain")= Q_("loss") |
|
Answer» ASSERTION andReasonarecorrectand Reasoniscorrectexplanationof Assertion |
|
| 39141. |
The gravitational potential energy of a rocket of mass 200kg at a height of 10^(7)m from the earth's surface is 10^(9)s. The potential energy of the rocket at a height of 26.4xx10^(6)m from the surface of earth is: |
|
Answer» `-1.5xx10^(9)J` `(U')/(U)=((R+h)/(R+h'))` `=((6.4+10)xx10^(6))/((6.4+26.4)xx10^(6))` `=(16.4)/(32.8)=(1)/(2)` `thereforeU'=-(U)/(2)=(-10^9)/(2)=-0.5xx10^(9)J` |
|
| 39142. |
If a drop of liquid breaks into smaller droplets, it results in lowering of temperature of the droplets .Let a drop of radius R, break into N small droplets each of radius r. Estimate the drop in temperature. |
|
Answer» Solution :Radius of larger drop =R Radius of smaller drop =r Volume of large drop =Volume of small drops N `(4)/(3)piR^(3)=N((4)/(3)pir^(3))` `thereforeR^(3)=Nr^(3)` `N=(R^(3))/(r^(3))`….(1) Change in area , `DeltaA` = area of large drop - area of small drops N `=4piR^(2)-N(4pir^(2))` `=4Deltapi(R^(2)-Nr^(2))` `therefore` Energy relased `=SDeltaA` `=S4pi(R^(2)-Nr^(2))`....(2) (Where S =SURFACE tension) Due to this energy relased ,temperature decrease in temperature is `DeltaQ`. Then energy RELEASED, `Q=CmDeltatheta` `=C((4)/(3)piR^(3)rho)Deltatheta` ...(3) (Where m = volume V `xx` density `rho=(4)/(3)piR^(3)rho)` Equating equation (1) and (2),`Sxx4pi(R^(2)-Nr^(2))=((4)/(3)piR^(3)rho)CxxDeltatheta` `thereforeDeltatheta=(Sxx4pi(R^(2)-Nr^(2)))/(((4)/(3)piR^(3)rho)C)` `Deltatheta=(3S)/(rhoC)[(R^(2)-Nr^(2))/(R^(3))]` `Deltatheta=(3S)/(rhoC)[(1)/(R)-(Nr^(2))/(R^(3))]` PUTTING the VALUE of N from equation (1), `Deltatheta=(3S)/(rhoC)[(1)/(R)-(R^(3))/(r^(3))((r^(2))/(R^(3)))]` `=(3S)/(5C)[(1)/(R)-(1)/(r)]`....(4) decrease in temperature. |
|
| 39143. |
The number of signifigures in the numbers 4.8000 xx 10^4 and 48000.50 are respectively |
|
Answer» 5 and 6 |
|
| 39144. |
Infinite number of bodies, each of mass 2 kg are situated on x-axis at distances. 1m, 2m, 4m, 8m,…. respectively, from the origin. The resulting gravitational potential due to this system at the origin will be |
|
Answer» `-(4)/(3)G` |
|
| 39145. |
A vertical cylinder contains an ideal gas enclosed, at a pressure of 2 atmosphere, by means of a freely moving piston containing weights. The cylinder is placed on a lift which rises with an acceleration of g//3. Find the fractional change in volume of the gas. Does the gas gets compressed? |
|
Answer» `(1)/(3)`, YES |
|
| 39146. |
A particle is projected with a velocityof 20 m/s at an angle of 30^@ with a plane of inclination 30^@ with respect to the horizontal . Theparticle hits the inclined plane at an angle of 30^@ . Find the (a) time of impact, (b) the height of the point of impact from the horizontal plane passing through the point of projection. |
|
Answer» |
|
| 39147. |
In oldage arteries carrying blood in the human body become narrow resulting in an increase in the blood pressure, this follows from |
|
Answer» PASCAL's law |
|
| 39148. |
Mention the advantages of writing the final results in significant figures. |
| Answer» Solution :Significant figures GIVE the NUMBER of meaningful DIGITS in a number. It helps us to KNOW about the extent of uncertainty in a MEASUREMENT. | |
| 39149. |
A drop of water of volume V is pressed between the two glass plates so as to spread on area A. If T is the surface tension, the normal force required to separate the glass plates is : |
|
Answer» Solution :PRESSURE difference = `T/R=T/(t//2)=(2T)/t`, FORCE = `(2T)/TA` But V = At or t = `V/A,THEREFORE"Force"=(2TA^(2))/V` 
|
|
| 39150. |
An artifical satellite moving in a circular orbit around the earth has total (kinetic + potential ) energy -E. Then : |
|
Answer» Its POTENTIAL ENERGY is `-2E` |
|