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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.
| 15251. |
A train is moving towards east and a car is along north, both with speed. The passenger in the train is observed the car moving in whivh of the following direction ? |
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Answer» East - NORTH DIRECTION |
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| 15252. |
If vec(A)+vec(B)=vec(R ) and 2vec(A)+vec(B) is perpendicular to vec(B) then |
| Answer» ANSWER :A | |
| 15253. |
A particle moves along a straight line such that its displacement at any time t is given by s= (t^3- 6t^3 +3t+4) metres. The velocity when the acceleration is zero is |
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Answer» 3m/s |
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| 15254. |
A satellite is revolving in a circular orbit at a height .h from the earth.s surface (radius of earth R, h«R). The minimum increases in its orbital velocity required, so that the satellite could escape from the earth.s gravitational field, is close to : (Neglect the effect of atmosphere.) |
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Answer» `SQRT(GR)(sqrt(2)-1)` |
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| 15255. |
A uniform cylinder of height h and radius r is placed with its circular face on a rough in inclined plane and inclination of the plane to the horizontal is gradually increased. If mu is the coefficient of friction |
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Answer» The CYLINDER will slide before toppling if `mult(2r)/(h)` |
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| 15256. |
A) When a ball hits a wall (stationary) it returns back with same velocity, if collision is elastic. B) A ball hits a bus moving towards it. If the collision is perfect inelastic, they move together. C) Inelastic collision arises due to loss in K.E during collision |
| Answer» Answer :C | |
| 15257. |
When a spring is compressed by a distance .x., the potential energy stored is U_(1). When it is further compressed by a distance 2x, the increases in potential energy is U_(2). The ratio of U_(1) : U_(2) is |
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Answer» `1: 7` |
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| 15258. |
30 gms of water at 30°C is in a beaker. Which of the following, when added to water, will have greatest cooling effect? (Specific heat of copper= 0.1 cal gm°C) |
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Answer» 100 GM of water at `10^(@) C` |
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| 15259. |
A particle of mass 2gm is initially displaced through 2 cm and then released. The fricitonal force constant due to air on it is 12xx10^(-3)N//m The restoring force constant is 50xx10^(-3)N//m. If it is in oscillatroy motion, its time period is |
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Answer» `PI` SEC |
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| 15260. |
The bob of a simple pendulum is displaced from its equilibrium position 'O' to a position 'Q' which is at a height 'h' above 'O' and the bob is then released. Assuming the mass of the bob to be 'm' and time period of oscillation to be 2.0 sec, the tension in the string when the bob passes through 'O' is |
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Answer» `m(G + 2 pi^2 H)` |
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| 15261. |
Water from a tap emerges vertically down with an initial speed of 1.0ms^(-1). The cross sectional area of tap is 10xx 10^(-5)m^(2). Assume that the pressure is constant throughout the stream of water, and that the flow is a steady, the cross sectional area of the steam 0.15m below the tap is |
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Answer» `5.0xx10^(-4)m^(2)` |
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| 15262. |
According to Jurin's law, the graph of the diameter (d) vs height of the water column (h) of the capillary tube will be |
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Answer» Circular |
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| 15263. |
A glass plate breaks when a mass of 3 kg or more is kept on it. An iron piece is kept on that plate in a lift. If the upward acceleration of the lift if gradually increased to 245 cmcdot s^(-2), the plate breaks, what is the mass of the iron piece? |
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Answer» |
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| 15264. |
A student forgot Newton's formula for speed of sound but he knows there were speed (upsilon) pressure (p) and density (d) in the formula. He then starts using dimensional analysis method to find the actual relation. upsilon = kp^(x)d^(y). Where k is a dimensionless constant on the basis of above passage answer the following questions.The value of x is |
| Answer» Answer :B | |
| 15265. |
A mass m =8kg is attached to a spring as shown in figure and held in position so that the spring remains unstretched. The spring constant is 200 N/m. The mass m is then released and begins to undergo small oscillations. Find the maximum velocity of the mass (g =10m//s^2) |
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Answer» SOLUTION :Mean position will be at k. x = mg or `x= (mg)/(k) = (8xx10)/(200) =2/5 =0.4m` This is ALSO the amplitude of OSCILLATION i.e. , A=0.4 m Now `v_("max") = A omega = Asqrt(k/m) = (0.4) SQRT((200)/(8)) = 2m//s` |
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| 15266. |
(A): If temperature of an ideal gas is constant. Then change in internal energy is zero.(R ):For a cyclic path the change in internal is equal to the work done by the gas |
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Answer» Both (A) and(R ) are true and (R ) is the CORRECT explanation of (A) |
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| 15267. |
A student forgot Newton's formula for speed of sound but he knows there were speed (upsilon) pressure (p) and density (d) in the formula. He then starts using dimensional analysis method to find the actual relation. upsilon = kp^(x)d^(y). Where k is a dimensionless constant on the basis of above passage answer the following questions.If the density will increase the speed of sound will : |
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Answer» Increase |
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| 15268. |
Can an instrument be called precise without being accurate? Can it be accurate withoutbeing precise? |
| Answer» SOLUTION :YES. An INSTRUMENT can be PRECISE WITHOUT being accurate. But it cannot he accurate without precision. | |
| 15269. |
An unknown metal of mass 192 g heated to a temperature of 100" "^(@)C was immersed into a brass calorimeter of mass 128 g containing 240 g of water at a temperature of 8.4" "^(@)C. Calculate the specific heat of the unknown metal if water temperature stabilizes at 21.5" "^(@)C. (Specific heat of brass is "394" J Kg"^(-1)K^(-1)) |
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Answer» `"458 J Kg"^(-1)K^(-1)` `M_(m)xxC_(m)xx(theta_(2)-theta_(1))=(M_(w)C_(w)+M_(w)xxC_(w))(theta_(2)-theta_(1))`. .(1) Here `M_(m)=` mass of metal piece = 192 g = 0.192 g `M_(w)=` mass of water = 240 g = 0.240 kg `C_(m)=` specific heat of metal = ? `C_(w)=` specific heat of water `=4186" kg"^(-1)K^(-1)` `theta_(2)=` TEMP. of metal piece `=100" "^(@)C` `theta_(1)` = steady temp. of mixture `=21.5" "^(@)C` `theta=` initial temp. of calorimeter + water `8.4" "^(@)C` By SUBSTITUTING values in equation (1), `:.0.192xxC_(m)(100-21.5)=(0.128xx394+0.240xx4186)[21.5-8.4]` `0.192xxC_(m)xx78.5=(50.432+1004.64)xx13.1` `15.072C_(m)=1055.072xx13.1` `15.072C_(m)=13821.4432` `:.C_(m)=(13821.4432)/(15.072)` `=917.0" J kg"^(-1)K^(-1)` `=916" J kg"^(-1)K^(-1)` (nearest value) |
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| 15270. |
ABCDEFGH is hollow cube made of an insulator, Fig. Face ABCD has positive charge on it. Inside the cube, we have ionized hydrogen. The usual kinetic theory expression for pressure |
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Answer» will be valid |
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| 15271. |
Which law of Newton is called the law of equilibrium ? |
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Answer» Newton's FIRST law of MOTION |
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| 15272. |
The rear side of a truck is open and a box of 40 kg mass is placed 5 m away from the open end as shown in Fig.The coefficient of friction between the box and the surface below it is 0.15.On a straight road, the truck starts from rest and accelerates with 2 m s^(-2) .At what distance from the starting point does the box fall off the truck?(Ignore the size of the box). |
| Answer» Solution : ACCELERATION of the BOX DUE to friction` = mu g = 0.15 xx 10 = 1.5 m s^(-2)` .But the acceleration of the TRUCK is greater.The acceleration of the box relative to the truck is` 0.5 m s^(-2)`towards the rear end.The time taken for the box to fall off the truck ` = SQRT( (2 xx 5)/(0.5)) = sqrt20 s ` .During this time, the truck covers a distance ` = 1//2 xx 2 xx 20 = 20m ` | |
| 15273. |
The kinetic energy of a body is 'K'. If one-fourth of its mass is removed and velocity is doubled, its new kinetic energy is, |
| Answer» ANSWER :B | |
| 15274. |
A simple pendulam of length .l. carries a bob of mass .m.. The bob is released from horizontal position. The tension in the string at t he lowest point would be |
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Answer» mg |
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| 15275. |
These orbital speed of satellite revolving very close to earth |
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Answer» `sqrt((2GM)/R)` |
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| 15276. |
What kind of relation exists between kinetic energy (E_k)and orbital radius (r) of the satellite revolving around the earth? |
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Answer» `E_k prop R ` `(mv^2)/(r) = (GM_em)/r^2` `:. 1/2 mv^2 = (GM_em)/(2R)` `:. E_k = ((GM_em)/2).1/r` `:. E_k prop 1/r [ :. (GM_em)/2 ` is cosntant ] |
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| 15277. |
A boy ''A'' of mass 50kg climbs up a stair case in 10s. Another boy ''B'' of mass 60kg climbs up a same staircase in 15s. The ratio of their power developed by the boys ''A'' and ''B'' is |
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Answer» <P>`5/4` `"Work done by the boy" .B. (W_(2)=m_(2)gh=60gh` power develop by`A=p_(1)=w_(1)/t_(1)=(50gh)/10` power developed by B=is`p_(2)=(60gh)/(15)."then",p_(1)/p_(2)=((50gh)/(10))/((60gh)/(15))=5/4` |
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| 15278. |
A block of mass 'm' rests on a rough inclined plane making an angle 30^(@) with the horizontal. The coefficient of static friction between the block and the plane is 0.8. If force of friction on the block is 10N, the mass of the block is (g=10 m//s^(2)) |
| Answer» ANSWER :D | |
| 15279. |
A hoop of radius 2 m weights 100 kg. It rolls along a horizontal floor so that its centre of mass has a speed of 20 cm/s. How much work has to be done to stope it? |
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Answer» Solution :Angular velocity of centre of mass of hoop `omega=(v)/(r )=(0.20)/(2)` `therefore omega=0.1" RAD s"^(-1)` Suppose moment of inertia about an axis passing through its centre and PERPENDICULAR to its plane is I. `therefore I=Mr^(2)` (ring TYPE) = `100xx(2)^(2)` `=400kgm^(2)` Now INITIAL total KINETIC energy of hoop = rotational kinetic energy + translational kinetic energy `K_(0)=(1)/(2)Iomega^(2)+(1)/(2)mv^(2)` `=(1)/(2)xx400xx0.01+(1)/(2)xx100xx400xx10^(-4)` `=2+2=4J` From work energy theorem in `W=K-K_(0),K=0` `=-4J` Negative sign indicate that work is done on hoop. |
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| 15280. |
A car is travelling with linear velocity (v) on a circular road of radius (r). If it is increasing its speed at the rate of a (mt//sec^(2)), then the resultant acceleration will be |
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Answer» `((V^(2))/(r)+a)` |
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| 15281. |
A ball is connected to an ideal string of length l=1.0 m whose other endis fixed to the point A. ball is released from the point B such that it can move in the vertical plane. Height of the point A above the ground is more than the length of string.The velocity of the ball at the lowest point of its motion is |
| Answer» Answer :A | |
| 15283. |
What does it mean when a height of a barometer is falling ? |
| Answer» Solution :A falling of a height of a barometer indicates DECREASING AIR pressure due to increase in water VAPOUR in ATMOSPHERE . It has possibility of rain fall . | |
| 15284. |
A mass mis suspended from a wire of length L. cross-section A and Young.s modulus Y. It is pulled along the length of the wire and released. Derive an expression for its time period. |
| Answer» Solution :`Y=(F)/(A)xx(L)/(x) rArr (F)/(x)=(YA)/(l) rArr K=(F)/(x)=(YA)/(l) rArr T=2pi SQRT((m)/(k)), T=2pi sqrt((m)/((YA)/(l))) rArr T=2pi sqrt((ml)/(YA))` | |
| 15285. |
The angular velocity of a particle moving in a circle realative to the center of the circle is equal to omega. Find the angular velocity of the particle relative to a point on the circular path. |
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Answer» Solution :`omega_(B A) = (v_(B A)_|_)/(A B)` where `v_(B A)_|_ = v COS THETA "and" AB = AC cos theta` because `ABC` is a right - angled triangle. Then, `omega_(B A) = (v)/(A C) = (v)/(2 R)` Substituting `(v)/( R) omega_(B O)`, we have `omega_(B A) = (1)/(2) omega_(B O) (-(1)/(2) omega))` Alternative procedure : `phi = 2 theta` Then, `(d phi)/(d t) (- omega_(B O) = 2(d theta)/(d t)` where `(d theta)/( dt) = omega_(B A)` This gives `omega_(B O) = 2 omega_(B A)`.  .
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| 15286. |
Find the average velocity for the time intervals Deltat=t_(2)-0.75 when t_(2) is 1.75,1.5,1.25and 1.0s . What is the instantaneous velocity at t=0.75 s? |
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Answer» |
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| 15287. |
A nail is located below the point of suspensions of a simple pendlum of length .l.. The bob is released from horizontal position. If the bob loops a verticle circle with nail as centre, find the distance of nail from point of suspension. |
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Answer» SOLUTION :Form the principal of conservation of MECHANICAL energy PE at the HORIZONTAL position = KE at the lowest position. `mgl = (1)/(2) mv^(2)` To loop the verticle circle, `v^(2) = 5gh (1-x)` where .x. is the distance of nail from point of suspension, and .v. is the velocity at lowest point of path `(mgl=(1)/(2)m.5g(l-x)),l=(5)/(2)(l-x), "" x = 3l//5` |
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| 15288. |
A shell is projected from a level ground with a velocity of 20m/s at 45° to the horizontal. When the shell is at the highest point it breaks into two equal fragments. One of the fragments whose initial velocity after the explosion is zero falls vertically downward. At what distance from the point of projection does the other fragment fall? (g=10 ms^(-2)) |
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Answer» 30 m |
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| 15289. |
The given P - V diagram is showing an ideal gas undergoing a change of state different paths I, II, III and IV that lead to the same change of state, then change in internal energy is |
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Answer» maximum along PATH 1 |
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| 15290. |
A gas expands from 1 litre to 3 litres atatmospheric pressure. The work done by the gas is about |
| Answer» ANSWER :B | |
| 15291. |
(A): At absolute zero, the substance hasinternal energy.(R) : At OK the substance posses thevibrational motion of atoms with in themolecule |
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Answer» If both (A) and (R) are true and (R) is thecorrect explanation of (A) |
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| 15292. |
Calculate the work done by a car against gravity in moving along a straighthorizontalroad . The mass of the car is 400 kg and the distance moved is 2 m . |
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Answer» Solution :Force of gravity acts on the car vertically DOWNWARD while car is MOVING along horizontal ROAD . i.e ., ANGLE between them is `90^(@)` `W =Fd cos theta` `= Fd cos 90^(@)` = 0 ` "" (:. d bot F = mg)` Work done by the car against gravity `W Fd cos 90^(@) = 0 "" ( :. cos90^(@) = 0 )` |
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| 15293. |
A linear harmonic oscillator of force constant 2xx10^6 Nm^(-1) and amplitude 0.01 m has a total mechanical energy of 160 J. Then find maximum P.E and maximum KE? |
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Answer» Solution :Maximum elasticP.E =`1/2 KA^2 = 1/2 xx 2 xx 10^6 xx (0.01)^2 = 100 J` As the oscillator goes from the EXTREME to mean position, this is CONVERTED into K.E. So, maximum K.E is 100 J. Since total energy is 160 J, maximum P.E is 160J. From this it is understood that at the mean position potential energy of the simple harmonic oscillator is MINIMUM which NEED not be zero. |
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| 15294. |
The average energy consumed by a human being in a day |
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Answer» 2400 cal `(10^7 J)/(4.2 xx 10^(3) J//kcal)~~ 2400 kcal.` |
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| 15295. |
If the escape velocity on earth is 11.2kms^(-1) its value for a planet having double the radius and 8 times the mass of the earth is |
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Answer» `10.2 KMS^(-1)` |
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| 15296. |
When two soap bubbles of radii r_1 and r_2 (r_2 lt r_1) adjoin, the radius of curvature of the common surface is |
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Answer» `r_2-r_1` `p_1-p_0=(4T)/(r_1) [ T=` surface TENSION of soap solution] In case of the second bubble, `p_2-p_0=(4T)/(r_2)` `therefore p_1-p_2=4T(1/r_1-1/r_2)` Again, if the radius of curvature of the COMMON surface is r, then , `p_1-p_2=(4T)/(r )` `therefore (4T)/(r )= 4T(1/r_1-1/r_2) or, 1/r=(r_2-r_1)/(r_1r_2) or, r=(r_1r_2)/(r_2-r_1)` |
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| 15297. |
Arrange the following number in the ascending order of thir order of magnitude(I) 7853(II) 0.187(III) 0.05 |
| Answer» Answer :A | |
| 15298. |
An open knife edge of mass 200 g is dropped from height 5m on a cardboard. If the knife edge penetrates distance 2m into the card board, the average resistance offered by the cardboard to the knife edge is (g = 10m//s^(2)) |
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Answer» 7 N Force due to air resistance = Ma |
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| 15299. |
Which of the following do (does) not depend on whether like rays are paraxial or not |
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Answer» (A) (C ) and (D) |
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| 15300. |
(A) : hat(i)+hat(j) is not a unit vector ( R) : The addition of unit vectors never gives unit vector |
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Answer» Both (A) and ( R) are ture and ( R) is the correct explanation of (A) |
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