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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.
| 15301. |
A film of soap solution is trapped between a vertical frame and a light wire ab of length 0.1m. If g = 10m//s^2. Then the load W that should be suspended from the wire to keep it in equilibrium is |
| Answer» ANSWER :D | |
| 15302. |
The radius of a large hemisphere whose" axis of symmetry is vertical is R = 50 cm. A small sphere of radius r= 5cmrolls without slipping with a speed of 2cm/s. The mass of the sphere is 100 gm. If it start's at the top from rest(i)what is the kinetic energy at the bottom? (ii) What fraction is rotational ? (iii) What fraction is translational? |
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Answer» Solution :Radius of the small sphere = `R = 0.5 m` Radius of the small sphere =r = 0.05 m Mass of small sphere m=0.100 kg Speed = v = 0.02 m/s TRANSLATIONAL KINETIC energy `=K_(r) =1/2 MV^(2)` ROTATIONAL kinetic energy `=K_(r) = 1/2 Iomega^(2)` `=1/2 xx (2/5 mr^(2)) xx v^(2)/r^(2) = 1/5 mv^(2)` Total K.E. of the small sphere `=K = K_(l) + K_(r) = 1/2 mv^(2) + 1/5 mv^(2)` `K = 7/10 mv^(2) = 7/10 xx 0.100 xx (0.02)^(2)` `K = 2.8 xx 10^(-5) J` (ii) `K_(r)/K =((1//5)mv^(2))/((7//10) mv^(2)) = 2/7` `(2/7)` of the total energy is rotational `K_(t)/K =((1/2)mv^(2))/((7//10)mv^(2)) = 5/7` `(5/7)` of the total energy is translational. |
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| 15303. |
Specific heat of oxygen at constant pressure is 0.2174 kcal/kg K. If the ratio of its specific heats is 1.4, find the universal gas constant given that J = 4200 J/kcal and molecular weight of oxygen is 32. |
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Answer» <P> Solution :We have SEEN that `C_(p)=(gammaR)/( gamma-1)``C_(p)=` Molar heat capacity at CONSTANT pressure = PRINCIPAL specific heat at constant pressure `xx` Molecular weight. A specific heat of oxygenat constant pressure is GIVEN in kcal/kg K. we have to convert it into J/kg K by multiplying with `J=4200` J/Kcal. `C_(p) = 0.2174 xx4200 xx 32` J/kmole `K= (gammaR)/( gamma-1) = (1.4R)/((1.4-1))` `:.R=(0.2174xx4200xx32xx0.4)/(1.4) = 8348`J/kmole K. |
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| 15304. |
A solid shere is rotating with an angular velocity omega. It is of moment is inertia 'I' about its diametrical axis and is of linear expansion coefficient alpha. When temperature is increased by DeltaT, the fractional change in its angular velocity is |
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Answer» `2alphaDeltaT` |
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| 15305. |
Four small lead balls of volumes 1cm^(3), 2cm^(3), 3cm^(3) and 4cm^(3) are descending down through the same liquid a) viscous force is more on 4(cm)^(3) ball b) terminal velocity is more 4(cm)^(3) ball c) viscous force is more in 1(cm)^(3) ball d) terminal velocity is more in 1(cm)^(3) ball |
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Answer» a, B are correct |
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| 15306. |
618 km//s is the escape velocity from the surface of |
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Answer» Earth |
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| 15307. |
An object weight 72 N the Earth. What its weight at a height R/2 from Earth. |
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Answer» Solution :We KNOW that`G = GM` ` g. = g (R/(R+h))^(2) = g(R/(R+R/2))^(2) = g ((2R)/(3R))^(2)` `g . = 4/9 g ` WEIGHT W. = `4/9 W = 4/9 (72)` |
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| 15308. |
An aeroplane moves upward for 500km, then moves 1000km in direction 37^(@) south of East and finally 100 sqrt2km moves in North -East direction. Mark correct option |
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Answer» The DISTANCE travelled by aeroplane is`(1500 + 100 sqrt2)km` |
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| 15309. |
In a compound microscope, the intermediate image is |
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Answer» VIRTUAL, ERECT and magnified |
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| 15310. |
Sand grains are being dropped vertically at a rate of 2 kgcdot s^(-1) on a conveyer belt moving at a constant velocity of0.1 mcdot s^(-1). Find the additional force needed to keep the belt moving. |
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| 15311. |
Mercury themometers can be used to measure temperature up to ……………… . |
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Answer» `260^(@)C` |
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| 15312. |
A smooth massless string passes over a smooth fixed pulley. Two masses m_(1) and m_(2) (m_(1) gt m_(2)) are tied at the two ends of the string. The masses are allowed to move under gravity starting from rest. The total external force acting on the two masses is |
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Answer» `(m_(1) + m_(2))G` `m_(1)g -T = m_(1) a"" `(1) ` T - m_(2)g = m_(2)a""`…(2) Now from equations (1) and (2), we get, a = `((m_(1) - m_(2)))/(m_(1) + m_(2))g ` Therefore, TOTAL external force ` = (m_(1) + m_(2)) a = (m_(1) - m_(2))g` The option C is correct. 
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| 15313. |
A 15Kg mass is accelerated from rest with a force of 100N. As it moves faster, friction and air resistance create an oppositely directed retarding force given by F_R = A + BV. where A = 25 N andB = 0.5 (N)/(m//s)At what velocity does the acceleration equal to one half of the initial acceleration ? |
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Answer» `25 ms^(-1)` |
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| 15314. |
Twobodies A and B of masses 5 kg and 10 kg in contact with each other rest on a table against a rigid wall . The coefficient of friction between the bodies and the table is 0.15.A force of 200 N is applied horizontally to A.What are (a) the reaction of the partition (b) the action-reaction forces between A and B ?What happens when the wall is removed? Does the answer to (b) change, when the bodies are in motion? Ignore the difference between mu_s and mu_k |
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Answer» Solution :We assume perfect CONTACT between bodies A and B and the rigid partition.In that case, the self-adjusting normal force on B by the partition (REACTION) equals 200 N. There is no IMPENDING motion and no friction.The action-reaction forces between A and B are ALSO 200 N. Whenthe partition is REMOVED, kinetic friction comes into play. Acceleration of `A+B = [ 200 - (150 xx 0.15 )]//15 = 11.8 ms^(-2)` Friction on A` = 0.15 xx 50 = 7.5 N` `200 - 7.5- F_(AB) = 5 xx 11.8` `F_(AB) = 1.3 xx 10^2 N` , opposite to motion `F_(BA) = 1.3 xx 10^2 N` , in the direction of motion |
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| 15315. |
When will the relative velocity of two moving cars be greater than their individual velocities? |
| Answer» SOLUTION :When both are MOVING in OPPOSITE DIRECTIONS. | |
| 15316. |
Acceleration of block 'm' is (theta lt 45^(@)) |
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Answer» `G SIN THETA` |
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| 15317. |
A projectiles given an initial velocity of (hati+2hatj)ms^(-1), when hati is along the ground and hatj is along the vertical. If g=10ms^(-2) find the equation of trajectory. |
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| 15318. |
At what temperature is rms velocity of hydrogen molecule equal to that of an oxygen molecule at 47^@C |
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Answer» SOLUTION :`v_(RMS)=SQRT((3RT)/(M))` Now, `{:("rms velocity of "),(H_2"molecule"):}}={:{("rms velocity of "),(O_2"molecule"):}` (or) `sqrt((3RxxT)/(2))=sqrt((3Rxx(47+273))/32)` (or) TEMPERATURE `T = (2xx320)/32=20K` |
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| 15319. |
A solid cylinder of mass 8 kg and radius 50 cm is rolling down a plane inclined at an angle of 30^(@) with the horizontal. Calculate (i) force of friction, (ii) aaceleration with which the cylinder rolls down and (iii) the minimum value of corfficient of friction so that cylinder does not slip while rolling down the plane. |
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| 15320. |
A body of mass 5 kg, initially at rest, moves under the action of an applied force of 15N on a horizontal surface with a coefficient of friction 0.25. Calculate (a) Work done by the force (b) Work done by friction (c ) Work done by net force (d) Change in kinetic energy in 5s. |
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Answer» Solution :GIVEN, `m=5kg,t=5s`, FORCE of FRICTION `f=mumg=0.25xx5xx9.8=12.25N` Net force on the body `=15-12.25=2.75N` ACCELERATION of body = `(2.75)/(5)=0.55ms^(-2)` 
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| 15321. |
Saturated water vapour is enclosed in a cylinder undr a piston and occupiesvolume V_(0)=5 litre at temperature t=100^(@)C. Find the mass of water formed after the volume under the piston is decreased isthermally to V=2 litres. Saturated vapour pressure at 100^(@)C =10^(5)Nm^(-2). |
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| 15322. |
A vessel of length l, breadth b and height h is filled completely with a liquid of density d (see figure) Calculate the thrust on each surface of the cube. |
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Answer» SOLUTION :Thrust = Pressure `xx ` Area Thrust on the base= hdg` xx ` lb LATERAL thrust on a VERTICAL wall with l and h as its sides = Pressure at the centre of gravity ` xx `area ` = ( ( 0+ h)/( 2)) DG xx lh ` ` "" = (1)/(2) h^(2)ldg. ` Lateral thrust on bh wall ` ""= (h)/(2)dg xx bh` `"" = (1)/(2)h^(2)b dg.` |
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| 15323. |
A long piece of paper is being pulled on a horizontal surface with a constant velocity upsilon along its length. Width of the paper is L. A small block moving horizontally, perpendicular to the direction of motion of the paper, with velocity v slides onto the paper. The coefficient of friction between the block and the paper is mu. Find maximum value of v such that the block does not cross the opposite edge of the paper. |
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| 15324. |
The mass of the moon is (1/81) of the earth, but the gravitational pull is (1/6) of the earth. It is due to the fact. |
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Answer» The RADIUS of MOON is (81/6) of the earth |
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| 15325. |
A particle of mass m kg is displaced from one given point to another given point under the action of several conservative and non conservative forces (Neglect relativistic considerations) Now match the following . {:(,"Column-1",," Column-2"),("(A)","Displacement of particle",,"(P) Path dependent"),("(B)","Work done conservative force",,"(Q) Path independent"),("(C)","Work done by non-conservative force",,"(R) Frame dependent"),("(D)","Angular displacement",,"(S) Frame independent"),(,,,"(T) Dependent on location of observer in a given frame"):} |
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| 15326. |
How does the energy of a simple harmonic oscillator depend on the amplitude ? |
| Answer» SOLUTION :The energy of the simple harmonic OSCILLATOR is DIRECTLY proportional to the square of the amplitude `(E PROP A^2)`. | |
| 15327. |
Which of the following group have different dimension |
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Answer» Potential DIFFERENCE, EMF, voltage |
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| 15328. |
What should be the lengths of steel and copper rod so that the length of steel rod is 10 cm longer than the copper rod at all the temperatures. Coefficients of linear expansion for copper and steel are alpha_(Cu)=1.7xx10^(-5) " "^(0)C^(-1) and alpha_("steel") = 1.1xx10^(-5) ""^(@)C^(-1) |
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Answer» Solution :t is given that the difference in length of the two rods is always 10 cm. Thus the EXPANSION in boththe rodsmust be same for all temperaturedifferences. Thus we can say that at all TEMPERATURE differences, we have `DeltaL_(Cu) =DeltaL_("steel") (or) alpha_(Cu) l_(1) DELTAT=Delta_(st) l_(2)DeltaT` [If `l_(1) and l_(2)` are the initial lengths of Cu and steel rods] (or) `alpha_(Cu)l_(1)=alpha_(st)l_(2)(or) 1.7l_(1)=1.1l_(2)""...(a)` It is given that`l_(2)-l_(1)=10cm "" ....(b)` From EQUATION (a) and (b) we have`((1.7)/(1.1)-1)l_(1)=10cm (or) l_(1)=(10xx11)/(0.6)=18.3cm` Now from equation(b) `l_(2)=28.3cm` |
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| 15329. |
A 70 kg man stands in contact against the wall of a cylindrical drum of radius 3 m rotating about its vertical axis with 200 rev/min. The coefficient of friction between the wall and his clothing is 0.15. What is the minimum rotational speed of the cylinder to enable the man to remain stuck to the wall (without falling) when the floor is suddenly removed ? |
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| 15330. |
A water hose 2 cm in diameter is used to fill a 20 litre bucket. If it takes 1 minute to fill the bucket, then the speed v at which the water leaves the hose is, |
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Answer» 150 cm/s |
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| 15331. |
A thermos bottle containing coffee is vigorously shaken. If coffee is considered a system, (a) does its temperature rise? |
| Answer» SOLUTION :Yes. There will be a slight INCREASE in the temperature because of the EXTERNAL work DONE on it. | |
| 15332. |
A particle of mass m strikes a horizontal smooth floor with a velocity u making an angle 'theta' with the floor and rebound with velocity v making an angle 'phi' with the floor. The coefficient of restitution between the particle and the floor is e. Then |
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Answer» the IMPULSE delivered by the floor to the body is mu(1+e) sin `THETA` |
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| 15333. |
Show that Reynold.s number R=(rhov_(0)d)/(eta) is dimensitons (rho is density fo the fluid, v_(0) is the critical velocity, d is diameter through which the fluid is flowing and eta is the coefficient of viscosity of the fluid). |
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Answer» Solution :`R=([ML^(-3)][LT^(-1)][L])/[(ML^(-1)T^(-1)])=[(ML^(-1)T^(-1))/(ML^(-1)T^(-1))]=[M^(0)L^(0)T^(0)]` `:.R` is DIMENSIONLESS. |
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| 15334. |
A motor car of mass 1000 kg runs over a bridge at 48 km/hr the roadway in the form of an arc of radius 20 m. Find the reaction between the car and the road at the lowest point of the car. |
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| 15335. |
A light cylindrical vessel is kept on a horizontal surface. Its base area is A.A hole of cross sectional area 'a' is made just at its bottom side. The minimum coefficient of friction necessary for sliding of the vessel due to the impact force of the emerging liquid is n xx 10 ^(-3). Find (a lt lt A). Given (a)/(A) = 2 xx 10 ^(-3) |
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| 15336. |
Two blocks of mass m_(1) = 3 kg and m_(2) = (1)/(sqrt(3)) kg are connected by a light inextensible string which passes over a smooth peg. The peg is fixed on the top of the wedge. The planes of the wedge supporting m_(1) and m_(2) are inclined at 30^(@) and 60^(@) respectively with the horizontal. Calculate the acceleration of the masses . |
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| 15337. |
Prove that in case of oblique elastic collision of two particles of equal mass if one is at rest, the recoiling particles always move off at right angles to each other |
Answer» Solution : In elastic COLLISION momentum is conserved. So, conservation of momentum along x-axis yields. `m u= mv_(1) cos theta_(1) + mv_(2) cos theta_(2)` i.e., `u= v_(1) cos theta_(1) + v_(2) cos theta_(2)`….(1) and along y-axis yields `0=v_(1) SIN theta_(1) - v_(2) sin theta_(2)` ....(2) Squaring and adding EQNS (1) and (2), we get `u^(2) = v_(1)^(2) + v_(2)^(2) + 2v_(1) v_(2) cos (theta_(1) + theta_(2))`....(3) As the collisions is elastic `(1)/(2) m u^(2) = (1)/(2) mv_(1)^(2) + (1)/(2) mv_(2)^(2)` i.e., `u^(2)- v_(1)^(2) + v_(2)^(2)`....(4) Put `u^(2)= v_(1)^(2) + v_(2)^(2)` in Eqn (3). we get `2v_(1) v_(2) cos (theta_(1) + theta_(2))=0` As it is given that `v_(1) ne 0 and v_(2) ne 0` so `cos (theta_(1) + theta_(2))= 0, "i.e.," (theta_(1) + theta_(2))= 90^(@)` |
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| 15338. |
A piece of copper having a rectangular cross section of 15.2 mm x 19.1 mm is pulled in tension with 44,500 N force, producing only elastic deformation. Calculate the resulting strain.Y_(Cu)=1.2 xx 10 ^(11) N//m^(2). |
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Answer» Solution :`Y= ("LONGITUDINAL stress")/("Longitudianal STRAIN")` `THEREFORE` ongitudinal strain `= ("Stress")/(Y) = (F)/(AY) = (F)/(l xx BXX Y)` `= (44500)/(15.2 xx 10 ^(-3) xx 19.1 xx 10 ^(-3) xx 1.2 xx 10^( + 11))=0.00127` `~~ 1.27 xx 10 ^(-3)` |
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| 15339. |
If the radius of moon is 1.7 xx 10^(6) m and its mass is 7.34 xx 10^(22) kg. Then its escape velocity is |
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Answer» `2.4 xx 10^(3)MS^(-1)` |
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| 15340. |
A particle is moving along a vertical of radiusr=20 m with a constant vertical circle of radius r=20 m with a constnt speed v=31.4 m/s as shown in figure. Straight line ABC is horizontal and passes through the centre of the circle.A shell is fired from point A at the instant when the particle is at C. If distance AB is 20sqrt(3) m and the shell collide with the particle at B, then prove tan theta=((2n-1)^(2))/(sqrt(3)). Where n is an integer.Further show that smallest value of theta is 30^(@) |
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Answer» Solution :As at the time of firing of the shell, the particle was at C and the shell collides with it at B, therefore thenumber of the revolutions completed by the particle is odd multiple of half i.e. `(2n-1)//2`, where n is an integer. Let T be the time period of the particle then `T=(2pir)/v=(2xx3.14xx20)/31.4=4` second If t be the time of the flight of the shell, then t=time of `[(2n-1)//2]` revolutions of the particle `=((2n-1))/2xx4=(2n-1)` second for a projectile the time of flight is GIVEN by `t=(2u sin theta)/g` Hence `(2usin theta)/g=(2(2n-1)`.......i The RANGE of the projectile is given by `R=(U^(2)sin 2 theta)/g` Hence `(u^(2) sin 2 theta)/g = 20 sqrt(3)` ......ii From equations i and ii `tan theta=((2n-1)^(2))/(sqrt(3))` For `theta` to be SMALLEST n=1 so `tan theta=((2n-1)^(2))/(sqrt(3))` |
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| 15341. |
In non-uniform circular motion, the resultant acceleration makes an angle with the radius vector is |
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Answer» `TAN^(-1)((ra_(t))/(v^(2)))` |
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| 15342. |
A small steel ball bounces on a steel plate held horizontally. On each bounce the speed of the ball arriving at the plate is reduced by a factor e (coefficient of restitution) in the rebound , so that V_("upward")=eV_("downward") If the ball is initially dropped from a height of 0.4m above the plate and if 10s later the bouncing ceases, the value of e is |
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Answer» `sqrt(2/7)` `t=sqrt((2h)/g)[(1+E)/(1-e)]` `THEREFORE 10=sqrt((2xx0.4)/(10))[(1+e)/(1-e)] ["taking" g=10m//s^2]` `or, e=(25sqrt(2)-1)/(25sqrt(2)+1)~~(17)/(18)` |
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| 15343. |
A man drops an apple in the lift. He finds that the apple remains stationary and does not fall. The lift is |
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Answer» GOING down with CONSTANT SPEED |
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| 15344. |
The potential energy of a projectile at maximum height is 3/4 times kinetic energy of projection. Its angle of projection is |
| Answer» Answer :C | |
| 15345. |
A body cools in 7 minute from 60^(0)C to 40^(0)C. Temperature after the next 7 minutes is 7x. Find .x.. (The temperature (in degree) of surrounding is 10^(0)C). |
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| 15346. |
Milk is poured into a cup of tea and is mixed with a spoon. Is this an example of a reversible process ? Give reason in support of your answer. |
| Answer» Solution :The work DONE during mixing is converted into HEAT of the MILK, which we cannot get back. So the PROCESS is an irreversible process. | |
| 15347. |
A sphere P of mass m and velocity v_i undergoes an oblique and perfectly elastic collision with an identical sphere Q initially at rest. The angle theta between the velocities of the spheres after the collision shall be |
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Answer» 0 `mvecv_i+mxx0=mvecv_(Pf)=mvecv_(Qf)` where `vecv_(Pf) and vecv_(Qf)` are the final VELOCITIES of SPHERES P and Q after collision respectively. `vecv_i=vecv_(Pf)+vev_(Qf)` `(vecv_i.vecv_i)=(vecv_(Pf)+vecv_(Qf)).(vecv_(Pf)+vecv_(Qf))` `vecv_(Pf). vecv_(Pf)+vecv_(Qf). vecv_(Qf)+2vecv_(Pf). vecv_(Qf)` or `v_i^2=v_(Pf)^2+v_(Qf)^2 + 2v_(Pf) v_(Qf) cos theta`...(i) According to conservation of kinetic energy , we get `1/2mv_i^2 =1/2mv_(Pf)^2 + 1/2mv_(Qf)^2 implies v_i^2=v_(Pf)^2 +v_(Qf)^2`...(ii) Comparing (i) and (ii), we get `2v_(Pf)v_(qf)cos theta=0 implies cos theta =0 or theta=90^@` |
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| 15348. |
A student measures the thickness of a human hair by looking at it through a microscope of magnification 100. He makes 20 observations and finds that the average width of the hair in the field of view of the microscope is 3.5 mm. What is the estimate on the thickness of hair ? |
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Answer» SOLUTION :Viewed WIDTH of human hair = 3.5 mm Magnification of microscope = 100 REAL width of human hair=? Magnification `=("viewed width")/("real width")` Real width `=("viewed width")/("magnification")` Real width `=(3.5)/(100)=3.5xx10^(-2) mm` |
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| 15349. |
Two bodies M and N of equal mass are hung separtely from two lightweight springs. Force constants of the springs are k_1 and k_2. The bodies are set to vibrate so that their maximum velocities are equal. Find the ratio of the amplitudes of vibration of the two bodies. |
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Answer» Solution :A body suspended from a SPRING will ACQUIRE maximum velocity at the equilibriumposition in its path of vibration. At that position potential energy of the body becomes ZERO and its energy becomes totally kinetic. Let the amplitude of vibration of the body M be `x_1` and that of N be `x_2`. Since they are of EQUAL MASS and their maximum velocities are also the same, their maximum kinetic energies will also be equal [`because` Maximum kinetic energy `=1/2timesmasstimes(maxim um velocity)^2`] . This kinetic energy transforms into the potential energy `(1/2kx^2)` of the body at the end of its amplitude. `therefore1/2k_1x_(1^2)=1/2k_2x_(2^2)or,x_1/x_2=sqrt(k_2/k_1)`. |
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| 15350. |
Determine the poisson's ratio of the material of a wire whose volume remains constant under an external normal stress. |
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Answer» Solution :Let l, D be the LENGTH and DIAMETER of the given wire RESPECTIVELY. As volume of wire remains constant. So`(piD^(2)l)/(4)=a"constant"""...(1)` Differentiating it, we have, `(pi)/(4)(2DDeltaDxxl+D^(2)Deltal)=0` DIVIDING it by `D^(2)l` we GET `(2DeltaD)/(D)+(Deltal)/(l)=0` (or)`(Deltal)/(l)=-(2DeltaD)/(D)` `sigma=(-DeltaD//D)/(Deltal//l)=(-DeltaD//D)/(-(2DeltaD//D))=(1)/(2)` `=0.5` |
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