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This section includes 7 InterviewSolutions, each offering curated multiple-choice questions to sharpen your Current Affairs knowledge and support exam preparation. Choose a topic below to get started.
| 1. |
A capacitor is to be designed to operate, with constant capacitance, in an environment of fluctuating temperature. As shown in the figure the capacitor is a paralel plate capacitor with 'spacer' to change the distance for compensation of temperature effect. If alpha_(1) be the coefficient of linear expansion of plates and alpha_(2) that of spacer, find the rate of change of capacitance with temperature and hence find the condition for no change in capacitance with change of temperature. The capacitance of the capacitor is equal to C. |
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| 2. |
A weightless rod of length l with a small load of mass m at one of its end is held vertical with its lower end hinged on a horizontal surface. The load touches a wedge of mass M in this position. A slight jerk towards right sets the system in motion (see figure), with rod rotating freely in vertical plane about its lower end. There is no friction. (a) For what mass ratio M/m will the rod form an angletheta=pi//3 with the vertical at the moment the load separates from the wedge?(b) What is speed of the wedge at that moment? Neglect friction. |
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| 3. |
The ends of an untagged rod are in contact with reservoirs maintained at 100^(@)C andat 0^(@)C through thin end-pieces of a material of lower thermal conductivity than lhe rod (Fig) which of the following graphs best represents the variation of temperature with distance between the fees of the reservoirs? |
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| 4. |
The length of a rod is measured as 31.52 cm. Granduations on the scale are upto |
| Answer» Answer :C | |
| 5. |
A 1m long rod having a constant cross sectional area is made of four materials. The first 0.2 m are made of iron, the next 0.3 m of lead, the following 0.2 m of aluminium, and the remaining part is made of copper. Find the centre of mass of the rod. The densities of iron, lead, aluminium, and copper are 7.9 xx 10^(3) kg//m^(3), 11.4 xx 10^(3), 2.7 xx 10^(3) kg//m^(3) and 8.9 xx 10^(3) kg//m^(3) respectively. |
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| 6. |
Three masses 700gm, 500gm and 400gm are suspended at the end of the spring and they are in equilibrium. When the 700gm mass is removed , the system oscillates with a oeriod of 3sec. When the 500gm mass is also removed , it will oscillate with a period of |
| Answer» Answer :B | |
| 7. |
Name the fundamental units and supplementary units on SI and their symbol of representation |
| Answer» SOLUTION :1 (B) 9 | |
| 8. |
State in the following cases, whether the motion is one, two or three dimensional: A kite flying on''a windy day. |
| Answer» SOLUTION :THREE DIMENSIONAL | |
| 9. |
A boat moves relative to water with a velocity which is .n. times the river flow (a)If n1 boat can cross the river along shortest path Boat can cross the river what ever is the value of n excluding zero value |
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Answer» only a is CORRECT |
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| 10. |
The velocity with which a projectile must be fired so that it escapes Earth's gravitational does not depend on........... |
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Answer» MASS of Earth |
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| 11. |
The internal energy of a system remainsconstant when it undergoes a) Cyclic process b) an Isothermal process c) Any process in which heat given out bythe system is equal to work done on thesystem |
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Answer» both a and B are true |
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| 12. |
The motion of a rocket is based on the principle of conservation of |
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Answer» mass |
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| 13. |
A piolt files due East from A to B. The velocity of the aeroplane in still air is vand the velocity of air relative to ground is u. The distance between A andB is l. Calculate the time for a round trip if air velocity is due to East. In still air time for round trip is t_(0) |
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Answer» `(t_(0))/(1-(U^(2))/(v^(2)))` |
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| 14. |
In Ohm's experiment to experiment to measure resistance different observation of unkowwn resistance are 4.12 Omega 4.08 Omega and 4.41 Omega. Find average absolute error, relative error and percentage error. |
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Answer» SOLUTION :ABSOLUTE ERROR `=0.16` RELATIVE error `0.0096`, PERCENTAGE error `=0.96%` |
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| 15. |
Using principle explain carburetor and spray pump. |
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Answer» Solution :The carburetor of automobile has a venturi channel (nozzle) through which air flows with a LARGE speed. The pressure is then lowered at the narrow neck and the petrol is sucked up in the chamber to provide the correct mixture of air to fuel necessary for combustion. Filter pumps , bunsenburner , atomisers and sprayers used for PERFUMES and sprayers usedfor insecticides work on the same principle (see FIGURE)  In SPRAY pump , piston forces air at HIGH speedscausing a lowering of pressure at the neck of the container and liquid is spray with air . |
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| 16. |
The energy required to shift a satellite from orbital radius .r. to radius 2r is E. What energy will be required to shift the satellite from orbital radius 2r to 3r ? |
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Answer» E |
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| 17. |
What is the dimensional formula and what is SI unit of coeffcient of thermal conductivity ? |
| Answer» SOLUTION :`MLT^(-3)THETA^(-1), Wm^(-1)K^(-1)` | |
| 18. |
According to Bernoullis theorem (p)/(d) + (v^(2))/(2) + gh= constant, dimensional formula of the constant is (p = pressure, d= density, h= height, v= velocity, g= acceleration due to gravity) |
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Answer» `M^(0)L^(0)T^(0)` |
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| 19. |
P is the centre of mass of a system of four point masses A, B, C and D, which are coplanar but not collinear. (a) P may or may not coincide with one of the point masses (b) P must lie within or on the edge of at least one of the triangles formed by taking A, B, C and D three at a time (c) P must lie on a line joining two of the points A, B, C, D (d) P lies outside the quadrangle ABCD |
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Answer» a & b are CORRECT |
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| 20. |
A steel tape 1 m long is correctly cabibrated for a temperature of 27.0^(0)C. The length of a steel rod measured by this tape is found to be 63.0 cm on a hot day when the temperature is 45.0^(0)C. What is the actual length of the steel rod on that day ? What is the length of the same steel rod on a day when the temperature is 27.0^(0)C ?Co-efficient of linear expansion of steel =1.20xx10^(-5)""^(@)C^(-1). |
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Answer» Solution :Length of the STEEL tape at `27^(0)C` is 100 cm. , i.e. `L=100cm and T =27^(0)C`. The length of steel tape at `45^(0)C` is `L.=L+DeltaL=L+alpha L DeltaT=100+(1.20xx10^(-5))xx100xx(45^(0)-27^(0))=100.0216cm` The length of 1 cm mark at `27^(0)C` on this scale, at `45^(0)C=100.0216//100` cm. Length of 63 cm MEASURED by this tape at `45^(0)C` will be `=(100.0216)/(100)xx63=63.0136cm` Length of the same steel rod on a DAY when the temperature is `27^(0)C=63xx1=63cm`. |
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| 21. |
Choose the correct pair of the following for this situation .During projectile motionthe quantities that remain unchanged are : |
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Answer» FORCE and vertical velocity |
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| 22. |
State and explain the work done in the following situations: A man spending his energy by pushing on a concrete wall. |
| Answer» Solution :Since DISPLACEMENT is ZERO, work DONE is zero. | |
| 23. |
A 10 kg stone is suspended with a rope of breaking strength 30 kg - wt. The minimum time in which the stone can be raised through a height 10 m starting from rest is (Take g = 10 N/kg). |
| Answer» ANSWER :B | |
| 24. |
Can a body have momentum without energy? |
| Answer» Solution :No, if a BODY has momentum, it must be in motion and CONSEQUENTLY POSSESS kinetic energy. | |
| 25. |
Dimensional formula of specific heat capacity is |
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Answer» `[ML^2T^-2K]` |
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| 26. |
The object reaches at maximum height of 20 m in 5 s when thrown upwards. Then what time will be taken by it to come to ground? |
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Answer» `2.5s` `therefore` Time taken to come to ground is 5 s. |
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| 27. |
A solid disc and a ring, both of radius 10 cm are placed on a horizontal table simulataneously, with initial angular speed equal to 10 pi rad s^(-1). Which of the two will starts to roll earlier ? The co - efficient of kinetic fricition is mu_(k)=0.2. |
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Answer» Solution :Frictional force acting on the body, `f = mu_(K)R = mu_(k)mg ""`…..(i) where R (= mg) is the normal reaction of the table on the body. If a is the acceleration of the CM of the body, `f = ma ""`….(ii) From eqns. (i) and (ii), `ma = mu_(k)mg` or `a = mu_(k)g ""`...(iii) Torque due to frictional force, i.e., `tau = fr = (mu_(k)mg)r=mu_(k)mgr` As `tau = I alpha.alpha=(tau)/(I)=(mu_(k)mgr)/(1)` Linear velocity of CM of the body (initially at rest, `upsilon_(0)=0`) is give by `upsilon = upsilon_(0)+at = at = mu kgt ""`....(iv) Angular velocity of the body after time t, i.e., `omega = omega_(0)+alpha t = omega_(0)-((mu_(k)mgr)/(I))t ""`......(v) (Here, `alpha` has been taken negative as torque due to frictional force produces retardation.) `upsilon = r omega ""`.....(vi) From eqns.(iv), (v) and (vi), we get `mu_(k)g t = r [omega_(0)-((mu_(k)mgr)/(I))t]=r omega_(0)-((mu_(k)mg r^(2))/(I))t ""`....(VII) For a RING, as `I = mr^(2)`, from eqn. (vii) `mu_(k) g t=r omega_(0)-((mu_(k)mgr^(2))/(mr^(2)))t=r omega_(0)-mu_(k)g t` or `2mu_(k)g t=r omega_(0)` or `t = (r omega_(0))/(2 mu_(k)g)` or `t = (0.1xx10pi)/(2xx0.2xx9.8)=0.80s` `("as " r = 0.1 omega_(0)=10 pi rad//s, mu_(k)=0.2, g=9.8 m//s^(2))` or `3 mu_(k)g t = r omega_(0)` or `t=(r omega_(0))/(3mu_(k)g)` or `t=(0.1xx10 pi)/(3xx0.2xx9.8)=0.53s` Since t is less in case of disc. the disc begin to roll earlier than the ring for the same value of r and `omega_(0)`. |
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| 28. |
A rod of weight W is supported by two parallel knife edges A and B and is in equilibrium in a horizontal position . The knives are at a distance d from each other . The centre of mass of the rod is at distance x from A . The normal reaction on A is |
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Answer» `(W(d- x))/(x)` `N_(1)` = Normal reaction on A `N_(2)` = Normal reaction on B W = WEIGHT of the rod  In vertical equilibrium `N_(1) + N_(2) = W "" ….. (i)` Torque BALANCE about centre of MASS of the rod , `N_(1) x = N_(2) (d - x)` Putting value of `N_2` from equation (i) `N_(1)x = (W - N_1) (d - x)` `implies N_1 x = Wd - Wx - N_1 d + N_1 x implies N_1 d = W (d - x)` `therefore N_(1) = (W (d - x))/(d)` |
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| 29. |
A block sliding down on a rough 45^(@) inclined plane has half the velocity it would have been, the inclined plane is smooth. The coefficient of sliding friction between the block and the inclined plane is |
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Answer» `(1)/(4)` `V_(R )=sqrt(2gl(sin theta-mu_(K)cos theta)), V_(S)=sqrt(2gl sin theta)` |
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| 30. |
“You cannot get something for nothing”. Do you think this statement is equivalent to any law of thermodynamics? |
| Answer» SOLUTION :YES, `I^("ST ")` LAW of THERMODYNAMICS. | |
| 31. |
In the above problem range of body is m. [g = 10 ms^(-2)] |
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Answer» `125sqrt(3)` |
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| 32. |
(A): Average velocity of gas molecules is zero. (R): Due to random motion of gas molecules, velocities of different molecules cancel cach other. |
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Answer» Both (A) and ( R) are TRUE and (R ) is the CORRECT explanation of (A) |
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| 33. |
Let T be the M.I. of uniform square plate about an axis AB that passes through its centre and is parallel to two of its sides. CD is a line in the plane of the plate that passes through the centre of the plate and makes an angle 9 with AB. The M.I . of the plate about the axis CD is equal to: |
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Answer» I |
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| 34. |
Prove that for a given velocity of projection, the horizontal range is same for two angles of projection alpha and (90^(@) - alpha). |
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Answer» Solution :The horizontal rangeis given by `R = (u^(2) sin 2 THETA)/( g) "". . . (1)` when `theta = alpha ` `R_(1)= (u^(2) sin 2 alpha)/( g) "". . . (2)` when`theta = (90^(@) - alpha)` `R_(2) = (u^(2) sin 2 (90^(@) - alpha))/( g)` ` = (u^(2) 2 sin (90^(@) - alpha) cos (90^(@) - alpha))/(g) "". . . (3)` But `sin (90^(@) - alpha) = cos alpha , cos (90^(@) - alpha) = sin alpha ` `R_(2) = (u^(2) (2 sin alpha cos alpha))/( g)` ` = (u^(2) sin 2 alpha)/( g) "" . . . (4)` From (2) and (4), it is SEEN that at both angles `alpha and (90 - alpha)`, the horizontal RANGE remains the same . |
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| 35. |
Bullets of 0.03 kg mass each hit a plate at the rate of 200 bullets per second with a velocity of 30 m//s. The average force acting on the plate in newton is |
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Answer» 120 |
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| 37. |
The radius in kilometers to which the presence radius of the earth (R = 6400 km) is to be compressed so that the escape velocity is increases to ten times is |
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Answer» 6.4 |
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| 38. |
A vessel of glass of constant volume contains 340g of mercury at 0^(@)C. When a few spherical iron balls are dropped into the vessel, 85g of mercury flow out. When the system is heated to 100^(@)C, 4.8g more of mercury flows out. Calculate the coefficeint of linear expansion of iron. (gamma_("mercury"=180xx10^(-6)K^(-1)) |
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| 39. |
A copper ballscools from 62^(@) to 50^(@)Cin 10 mintuesin the next10 minutes . Calculate its temperatureat the endof further 10 minutes . |
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Answer» Solution :Theball cools from `theta_(1) = 62^(@) C ` to ` theta_(2) = 50^(@)C ` in `t_(1) = 10`minutes . (i) Average temperature during the interval ` theta = (theta_(1) + theta_(2))/(2) = 56^(@)C` Average rate of cooling `(d theta)/(dt) = (theta_(1) - theta_(2))/(t_(1)) = (62 - 50)/(10) = 1.2^(@) C min^(-1)` `(d theta)/(dt) prop (theta - theta_(0)) theta_(0)` = temperature of the surroudungs `therefore1.2 prop (56- theta_(0))` (ii)The ball cools from `theta_(2) = 50^(@)C , theta_(3) = 42^(@)C , t_(2)= 10` mintues . `theta^(1)= ( theta_(2) + theta_(3))/(2) = 46^(@)C , (d theta^(1))/(dt)= (theta_(2) - theta_(3))/(t_(2)) = (50 - 42)/(10) = 0.8^(@)C min^(-1)` `(d theta^(1))/(dt) prop(theta^(1) - theta_(0)) therefore0.8 (46-theta_(0)) "".........(2)` Divinding(1) by(2) we get`(1.2)/(0.8) = (56-theta_(0))/(46 - theta_(0))thereforetheta_(0) = 26^(@)C` The BALLS cools from `therefore _(3) = 42^(@) to therefore _(4) ` in `t_(3) = 10 ` minutes`therefore theta= (theta_(3) + theta_(4))/(2) = (42 + theta_(4))/(10), (d theta)/(dt) = (theta_(3) - theta_(4))/(t_(3)) = (42 - theta_(4))/(10)` Now `(42-theta_(4))/(10) prop ((42 + theta_(4))/(4)) - 26 ""......(3)` DIVIDING eq. (3) by (2) we have . `(((42-theta_(4))/(10))/0.8) = ((42+(theta_(4))/(2) - 26))/(46-26) "" therefore(42 - theta_(4))/( 8) = (42 + theta_(4) - 52)/(2 xx 20)` `5(42 - theta_(4)) = theta_(4)- 10 "" therefore 6 theta_(4) = 220 "" therefore _(4) = 36.7^(@)C` |
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| 40. |
(A) : A hydrogen filled balloon stops raising after it has attained a certain height in the sky. (R ) : The buoyancy of an object depends both on the material and shape of the object. |
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Answer» Both (A) and (R ) are TRUE and (R ) is the correct EXPLANATION of (A) |
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| 41. |
A bus accelerates from rest at a constant rate alpha for some time, after which it decelerates at a constant rate beta to come to rest. If the total time elapsed is t seconds then, evaluate. (a) the maximum velocity achieved and (b) the total distance travelled graphically. |
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Answer» Solution :(a) Let `t_(1)` bethe time of acceleration and `t_(2)` that of deceleration of the bus. The TOTAL time is `t=t_(1)+t_(2)`. Let `v_(max)` be the maximum velocity. As the acceleration and deceleration are constant the velocity time GRAPH is a straight line as shown in the figure with `+ve` slope for acceleration and `-ve` slope for deceleration. From the graph, the slope of the line OA gives the acceleration `ALPHA`. `therefore alpha` = slpe of the line `OA=(v_(max))/(t_(1))impliest_(1)=(v_(max))/(alpha)` the slope of AB gives the deceleration `beta` `therefore beta` = slope of `AB=(v_(max))/(t_(2))impliest_(2)=(v_(max))/(beta)` `t=t_(1)+t_(2)=(v_(max))/(alpha)+(v_(max))/(beta)` `t=v_(max)((alpha+beta)/(alphabeta))` `therefore v_(max)=((alphabeta)/(alpha+beta))t` (b) DISPLACEMENT = area under the v-t graph = area of `DeltaOAB` `=(1)/(2)("base")("height")=(1)/(2)tv_("max")` `=(1)/(2)t((alphabetat)/(alpha+beta))=(1)/(2)((alphabetat^(2))/(alpha+beta))` 
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| 42. |
(A) : The minimum number of coplanar vectors whoser sum can be zero is three ( R) : The three vectors must be coplanar to produce equilibrium |
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Answer» Both (A) and ( R) are ture and ( R) is the correct explanation of (A) |
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| 43. |
In an orbital motion, the angular momentum vector is |
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Answer» ALONG the RADIUS vector |
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| 44. |
Explain the process of melting of ice with explanation of freezing, melting and melting point. |
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Answer» Solution :The change of STATE from solid to liquid is called melting and from liquid to solid is called fusion. It is observed that the temperature remains constant until the entire aount of the solid substance melts. Both the solid and liquid states of the substance coexist in thermal equilibrium during the change of states from solid to liquid. The temperature at which the solid and the liquid states of the substance is in thermal equilibrium with each other is called its melting point. It is characteristic of the substance. It also depends on pressure. The melting point of a substance at standard atmospheric pressure is called its normal melting point. An activity to explain the process of melting of ice :  Take a SLAB of ice. Take a metallic wire and fix two blocks, say 5 kg each, at its ends. Put the wire over the slab as shown in Fig. You will observe that the wire passes through the ice slab. This happens due to the fact that just below the wire, ice melts at lower temperature due to increase in pressure. When the wire has passed, WATER above the wire freezes again. Thus the wire passes through the slab and the slab does not split. This phenomenon of refreezing is called regelation. SKATING is possible on snow due to the formation of water below the skates. Water is FORMED due to the increase of pressure and it acts as a lubricant. |
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| 45. |
A lift is going up. The total mass of the lift and the passengers is 1500 kg. The variation in the speed of the lift is given by the graph as shown in figure. a) What will be the tension in the rope pulling the lift at time t equal to i) 1 sec ii) 6 sec iii) 11 sec? b) What is the height to which the lift lakes the passengers? c) What will be the average velocity and the average acceleration during the course of the entire motion? |
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| 46. |
A water in glass thermometer has density of water marked on its stem [Density of water is the thermometric property in this case]. When this thermometer is dipped in liquid A the density of water read is 0.99995 g cm^(3). Thereafter it is dipped in liquid B and the reading remains unchanged. Maximum density of water is 1.00000 g cm^(-3). (a) Can we say that liquid A and B are necessarily in thermal equilibrium? (b) If two liquids are mixed and the thermometer is inserted in the mixture, the height of water column in stem is found to change (i.e. reading is different from 0.99995 g cm^(-3)). Has the height increased or decreased? |
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| 47. |
How long will a boy willing near the window of a train travelling at 36kmh^(-1) sec a train passing by in the opposite direction with a speed of 18kmh^(-1). The length of slow-moving train is 90m. |
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Answer» Solution :The relative velocity of the slow-moving TRAIN with respect to the boy is `=(36+18)kmh^(-1)=54xx(5)/(18)ms^(-1)=15MS^(-1)`. SINCE the boy will watch the full length of the other train, to find the TIME taken to watch full train: We have, `15=(90)/(t)ort=(90)/(15)=6s` |
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| 48. |
A specific gravity bottle contains 50 xx 10 ^(-3)Kgmercury at 0^(@)С. Find the temperature at which 49.237 xx 10 ^(-3) Kg mercury completely fills it. (gamma _(R ) of mercury is 1.82 xx 10^(-4)//^(@)C and alpha of glass is 9 xx 10 ^(-6) //^(@)C) |
| Answer» SOLUTION :`99.97^(@)C` | |
| 49. |
The potential energy U between two molecules as a function of the distance X between them has been shown in the Fig. The two molecules are |
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Answer» ATTRACTED when X lies between A and B and are REPELLED when X lies between B&C |
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| 50. |
A body during its free fall from a height H, hits an inclined plane at a height h in its path. Due to collision,direction of the velocity of the body becomes horizontal. What should be the valueofh/H so that the body takes maximum time to reach the ground ? |
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