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101.

What is the reason for capillary rise?

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Answer :CAPILLARY RISE OCCURS when the adhesive force is GREATER than the cohensive force.
102.

Lift an object up. Work is done by the force exerted by you on the object. The object moves upwards. The force you exerted is in the direction of displacement. However, there is the force of gravity acting on the object. Which one of these forces is doing positive work? Give reason.

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Solution :The WORK done by the force applied by me is POSITIVE, because its acts in the DIRECTION of DISPLACMENT of the object.
103.

Lift an object up. Work is done by the force exerted by you on the object. The object moves upwards. The force you exerted is in the direction of displacement. However, there is the force of gravity acting on the object. Which one is doing negative work? Give reason.

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SOLUTION :The work done by the force of gravity is negative, because the force of gravity acts in the downward DIRECTION, i.e.,in the opposite direction of the DISPLACEMENT of the object.
104.

A gold ornament weighs 570 gram in air nad 520 gram in water. If the specific gravity of gold is 19, find the difference in the volume of water displaced when the ornament is immersed in water and the actual volume of the gold in the ornament. How do you account for this difference in volume ?

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Solution :Find the weight of gold ornament in air `(w_(1))` and water `(w_(2))`.
Find the density of gold ornament using formula,
`d=(w_(1))/(w_(1)-w_(2))`
Find the VOLUME of gold ornament using formula, `V=(m)/(d)=(w_(1))/(d)`
Is this equal to the apparent loss of weight of the gold ornament ?
Find the volume of 570 GRAMS of pure gold using formula, `d=(m)/(v)`
IS thereany difference in the volume of pure gold and gold ornament ?
Does the gold ornament CONTAIN a cavity ?
105.

Find out the reasons for the following. Though an egg sinks in pure water,it floats on salt water.

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ANSWER :As pure water has lower density it has LESS buoancy. Saline water has GREATER density and has greater density and has greater BUOYANCY. So an egg SINKER in pure water and floats on salt water.
106.

Name A, B, C, D, E and F in the following diagram showing change in its state.

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Solution :A - Melting POINT / lignation
B - VAPOURISATION
C - Condensation
D - Solidification
E - Sublimation
F - Sublimation
107.

Name the method used to identify different atomicand molecular content in vapours.

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108.

An object of mass 100g moves with speed 1ms^(-1). Its kinetic energy would be ………..J.

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50
5
`0.5`
`0.05`

Solution :`E_(K)=(1)/(2)MV^(2)`
`=(1)/(2)(100xx10^(-3))(1)^(2)`
`=50xx10^(-3)`
`=0.05J`
109.

in which of the following cases of motion, the distance covered and magnitude of displacement can be same?

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If the car performs motion on LINEAR path.
If the car performs motion on circular path.
If the oscillator OSCILLATES to and FRO about the mean-position.
If the earth REVOLVES around the sun.

Solution :Based on knowledge.
110.

What name is gives to those aircrafts which fly at speeds greater than the speed of sound?

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ANSWER :SUPERSONIC AIRCRAFTS
111.

A vernier callipers has its main scale graduated in mm and 10 divisions on its vernier scale are equal in length to 9 mm. When the two jaws are in contact, the zero of vernier scale is ahead of zero of main scale and 3rd division of vernier scale coincides with a main scale division. Find the zero error of vernier callipers.

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SOLUTION :`+ 0.03 ` CM
112.

A bat can hear sound of frequencies up to 120 kHz. Determine the minimum wavelength of sound which it can hear. Take speed of sound in air to be 344 ms^(-1)

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Solution :Given , F= 120 kHz = `120 xx 10^3 HZ, V= 344 m s^(-1)`
From RELATION `V = flambda`
WAVELENGTH `lambda = V/f = (344)/(120xx10^3)`
`=2.87xx10^(-3) m ` ( or 2.87 mm)
i.e. , the bat can hear sound of minimum wavelength 2.87 mm.
113.

The shape of an object helps to determine whether the object will float.

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ANSWER :1
114.

A certain household has consumed 250 units of energy during a month. How much energy is this in joule?

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Solution :1 UNIT of energy = `1kWh=3.6xx10^(6)J`
`THEREFORE` 250 units of energy = `250xx3.6xx10^(6)J`
`=900xx10^(6)J`
`=9xx10^(8)J`
115.

When a ray of light passes from air into glass, is the angle of refraction greater than or less than the angle of incidence?

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Solution :When a ray of light travels from rare (air) medium to DENSE [glass] medium, light ray MOVES TOWARDS the normal.
`therefore` Angle of refraction < Angle of INCIDENCE.
116.

Answer the following: How does the speed of sound differ in different media ?

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Solution :SPEED of sound is MAXIMUM in solids, LESS in liquids and least in gases.
117.

A wooden cube of side 10 cm has mass 700g What part of it remains above the water surface while floating vertically on the water surface?

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ANSWER :3CM HEIGHT
118.

A bulb draws current 1.5 A at 6.0 V. Find the resistance of filament of bulb while glowing.

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ANSWER :`4.0 OMEGA`
119.

Among the following ____ represents the smallest temperature change

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1 K
`1""^(@)C`
`1""^(@)F`
Both 1 and 2

Answer :A::C
120.

(a) What are infrasonic waves? Name two animals which produce infrasonic waves. (b) What are ultrasonic waves? Name two animals which can produce ultrasonic waves

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121.

Trip switch is an……….safety device.

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SOLUTION :ELECTROMECHANICAL
122.

Light ray emerges from water into air. Draw a ray diagram indicating the change in its path in water.

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SOLUTION :When a RAY of light travels from dense medium to rarer medium [ from WATER medium to air medium ], light ray moves away from the normal.
`THEREFORE` ANGLE pf incidence < Angle of refraction.
123.

M.A. is always greater than 1 in

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I CLASS levers
II class levers
III class levers
All the above

Answer :B
124.

Renewable source of energy is

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coal
fossil fuels
natural gas
sun

Answer :D
125.

The change of state from solid to gas directly is called _________.

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SOLUTION :MELTING POINT
126.

The velocity - time graph of an ascending passenger order lift is given in figure. What is the acceleration of the lift: (i) during the first two seconds , (ii) between 2nd and 10 th second, (iii) during the last two seconds. .

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SOLUTION :(i) 2.3 m/`s^(2)`(II) ZERO(III) -2.3 m/`s^(2)`
127.

Correct the following. Value of g remains the same on all regions of earth.

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Answer :VALUE of G is DIFFERENT on all REGIONS of earth.
128.

Which one among kinetic energy, potential energy and mechanical energy cannot be negative?

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KINETIC energy
potential energy
mechanical energy
Both 'B' and 'C'

ANSWER :C
129.

A lamp consumes 1000 J of electrical energy in 10s. What is its power?

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Solution :Here, `W=1000J, t=10S, P=?`
`P=(W)/(t)=(1000J)/(10s)=100Js^(-1)=100` watt
130.

In the previous example, calculate the mechanical advantage of the vehicle lifting machine.

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Solution :Given are of the pump PISTON, `a_(1)=0*25m^(2)`
area of the press piston, `a_(2)=5M^(2)`
MECHANICAL advantage of the machine,
`"M.A"=("Area of the press pison"(a_(2)))/("Area of the pump pison" (a_(1)))=(5)/(0*25)=20`
131.

Name one d.c. source and one a.c. source.

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ANSWER :d.e. SOURCE : CELL, a.c. source : MAINS.
132.

The magnitude of momentum of a movig object is equal to its kinetic energy. What will be the velocity of the object considering SI units for the system.

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SOLUTION :Since MOMENTUM (mv) = Kinetic ENERGY
`(1/2 mv^(2))`
= mv = `1/2 mv^(2)`
V= 2 m/s
133.

An elephant weighing 40,000 N stands on one foot of area 1000 cm^(2) whereas a girl weighing 400 N is standing on one 'stiletto heel of area 1 cm^2 (a) Which of the two, elephant or girl, exerts a (b) What pressure is exerted on the ground by the elephant standing (c) What pressure is exerted on the ground by the girl standing (d) Which of the two exerts larger pressure on the ground : elephant or girl? (e) What is the ratio of pressure exerted by the girl to the pressure exerted by the elephant ?

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Answer :Elephant has a larger weight of 40000 N, therefore, elephant exerts a larger force on the ground ; Elephant exerts a larger force on the ground by40000 N-400N 39600 N (b) `400,000 N//m^2 (C) 4000,000 N//m^2 `(d) Girl (E) 4000,000: 400,000 = 10: 1 The PRESSURE exerted by girl is 10 TIMES greater than that exerted by the elephant
134.

Ram is a college student in Delhi. Ram and his family are going by car to visit a hill station. Ram himself is driving the car. Ram drives the car very carefully. Before starting to drive, Ram has fastened the car seat belt himself properly. He has also made his father, mother and younger brother fasten their car seat belts. on the flat highwat road, Ram is keeping the car speed within a range of 50 to 60 kmph (which is well within the prescribed speed limit on this highway). He does nto accelerate the car unnecessarily. After driving for about five hours continuously on a flat road, there is a sight of hills in view. On approaching the hilly road, Ram increases the speed of his car. Ram's younger brother Anish, who is a student of class VI, is surprised to see his brother increasing the speed of car suddenly. Anish asks Ram why the speed of car has been increased. Ram explains the reason for increasing the speed of car to everyone. (a) What type of energy is possessed by the car while running on the flat road? (b) What type of energy transformation take place in a car engine? (c) When the car is moving on the flat road, it has to do work to overcome mainly two types of forces. Name these two types of forces. (d) When the car is moving on an uphill road, it has to do work to overcome three types of forces. Name these three types of forces. (e) Why does Ram increase the speed of his car on approaching the hilly road? (f) What types of energy is possessed by the car going up on the hilly road? (g) What values are displayed by Ram in this episode?

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Solution :(a) The car running on a flat ROAD possesses 'kinetic energy'.
(b) The car burns petrol as fuel. The car engine FIRST converts the chemical energy of petrol into heat energy. This heat energy is then converted into kinetic energy (which drives the wheels of the car). The transformation of energy taking place in a car engine can be written as:
`underset(("of petrol"))("Chemical energy")rarr"Heat energy"rarr"Kinetic energy"`
(C) When the car is moving on a flat road, it has to do work to overcome (i) friction of the road, and (ii) air RESISTANCE.
(d) When the car is moving on an uphill road, then it has to do work to overcome (i) friction of the road (ii) air resistance, and (iii) force of gravity.
(e) Ram increases the speed of car on APPROACHING a hilly road to give more kinetic energy to the car so that it may go up the hill aganist gravity.
(f) The car going up on the hilly road possesses (i) kinetic energy, and (ii) gravitational potential energy.
(g) The various values displayed by Ram in this episode are (i) Concern for the safety of his family (as shown by the fastening of car seat belts and driving within speed limit) (ii) Conservation of fuel or petrol (by driving the car within a specified speed range and avoiding unnecessary accelerating) (iii) Awareness (that car needs more kinetic energy to go up on a hilly road), and (iv) Knowledge (that kinetic energy of car can be increased by increasing its speed).
135.

An object floats in three immiscible liquids A, B and C of densities 3"g cm"^(-3),2"g cm"^(-3), respectively as shown in the figure. When the object is placed in the liquids, the levels of liquid A, B and C rise by 3 cm, 5 cm and 8 cm, respectively. The areas of cross-sections of the container and the object are 10 cm^(2)and5cm^(2), respectively. Calculate the density of the object.

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SOLUTION :WEIGHT of the floating body = Weight of the liquids A,B and C displaced.
VOLUME of the body immersed in a liquid = (Rise in level) `xx` (area of cross-section of the CONTAINER)
`V_(A)=(3cm)xx(10cm^(2))`
`V_(A)+V_(B)=(5cm)xx(10cm^(2))`
`V_(A)+V_(B)+V_(C)=(8cm)xx(10cm)`= Total volume of the body
Weight of A displaced `=V_(A)xxd_(A)`
Similarly, determine the weights of liquids B and C displaced.
Determine the density from the definition,
`"Density"=("mass")/("volume")`
136.

The pitch of a screw gauge is 1 mm and its circular scale has 100 divisions. In measurement of the diameter of a wire, the main scale reads 2 mm and 45th mark on circular scale coincides with the base line. Find the least count .

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SOLUTION :0.001 CM
137.

The pitch of a screw gauge is 1 mm and its circular scale has 100 divisions. In measurement of the diameter of a wire, the main scale reads 2 mm and 45th mark on circular scale coincides with the base line. Find the diameter of the wire.

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SOLUTION :0.245 CM
138.

A wooden block floats in water with two third of its volume submerged. A. Calculate the density of wood.b. When the same block is placed in oil, three - quarter of its volume in immersed in oil. Calculate the density of oil.

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ANSWER :a. `667kgm^(-3)` B. `889kgm^(-3)`
139.

Fill in the blanks The force required to produce an acceleration of1 m/s^2on a body of mass 1 kg is ………………..

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ANSWER :1 NEWTON (1N)
140.

A stone is dropped from the top of a tower 500m height into a pond of water at the base of the tower. When is the splash heard at the top ? Given g=10ms^(-2) and speed ofsound =340m//s.

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Solution :Height of the tower, `S=500m`.
Velocity of sound , `upsilon=340ms^(-1)`
Acceleration due to GRAVITY, `g=10m//s^(2)`.
Initial velocity of the stone, `u=o` (since the stone is initially at rest)
TIME TAKEN by the stone to fall to the baseof the tower, `t_(2)`
According to the second EQUATION of motion.
`S=ut_(1)+1//2 x 10x t_(1)^(2)`
`t_(1)^(2)=100 "" t_(1)=10S`
Now, time taken by the sound to reachthe TOP from the base of the tower, `t_(2)=500//340=1.47`s
Therefore, the splashis heard at the top after time, t
Where, `t=t_(1)+t_(2)=10+1.47=11.47S`.
141.

Anirregularsolidwhenimmersed in waterdisplaces 3/4th litre ofwater. Whenimmersedin agiven liquid it displaces 600 g of theliquid. What isthe densityof theliquid?

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Solution :(i) `Z .("PITCH")/("No.of C.C.D's")`
Dimeansionalformula of z
`=(("demensional forumla of y")^(2))/("dimensional FORMULA of x")`
(ii) `[M^(1)L^(0)T^(0)]`
142.

A wave is slinky travelled to and fro in 5 see the length of the slinky is 5m. The velocity of wave is_________.

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10m/s
5m/s
2m/s
25m/s

Answer :C
143.

A ball is initially moving with a velocity 0.5 m s^(-1) ms. Its velocity decreases at a rate of 0.05 m s^(-2). (a) How much time will it take to stop ? (b) How much distance will the ball travel before it stops ?

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SOLUTION :GIVEN, initial velocity u = 0.5 m `s^(-1)`, final velocity v = 0, acceleration a = - 0-05 m `s^(-2)`? (Here negative sign is used since velocity decreases with time).
(a) From equation of motion v = u + at
`0= 0.5 - 0.05 xxt`
or 0.05 t = 0.5
or `t = (0.5)/(0.05)` = 10 s
(b) From equation of motion `v^(2)= u^(2)+ 2A S `
`0 =(0.5)^(2)- 2 xx0.05 XXS `
or `0.1 S = 0.5 ` or S `= (0.25)/(0.1)= 2.5 ` m
144.

Feroz and his sister Sania go to school on their bicycle. Both of them start the at the same time from their home but take different times to reach the school although they follow the same route. The table given below shows the distance travelled by them in different times:

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SOLUTION :
145.

Why are the cellings of concert halls curved ?

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SOLUTION :CEILINGS of concert HALLS are curved so that sound after REFLECTION (from the walls) spreads uniformly in all directions.
146.

In a certain experiment , a sound wave was observed to have undergone a change in its velocity and wavelength but the frequency remained the same . In another experiment , no change was observed in the velocity , wavelength or frequency but there was a change in the phase . if the direction of the wave propagation is changed in both the cases , identify the phenomena that took place in the two experiments . Give reasons for the changes in the physical quantities .

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Solution :Velocity of a wave CHANGES whenever SOUND travels from one medium to ANOTHER medium . Change in the velocity results in refraction . When a sound wave is reflected by an accountably hard , smooth surface , velocity and wavelength (thus , frequency too) remain CONSTANT , but phase changes by `180^(@)` . Thus, the first experiment refers to refraction of sound waves and the second experiment refers to reflection of sound waves.
147.

The average mass of an atom of uranium is 3.9xx10^(-25) kg. Find the number of atom is 1 g of uranium.

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Solution :Given :mass of ONE atom of uranium
`= 3.9xx10^(-25) KG = 3.9xx10^(-22)` g
(SINCE 1 kg = `10^3` g)
`therefore ` Number of ATOMS in 1 g uranium
`= 1/(3.9xx10^(-22)) = 2.56xx10^(21)`
148.

An object and its image are as shown in the diagram below. If the object image distance is 4 cm and the magnification is 3, find type of thelens used and ratiou : v : f. Draw the ray diagram toshow the position of the lens and the principal focil.

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Solution :(i) As THEMAGNIFICATION in more than 1, lens MUST be convex lens.
(ii) MAGNIFICATION `m = v/u`(1)
v = u + d
where 'd' is thedistance between the image andtheobject.
Substitude the values of v in (1) anddetermins thevalues of u and v by using (1).
Now thefocal length 'f' may be determined asltBrgt ` f = (UV)/((u-v))`
FIND the ratioof u, v andf .
` 2 : 6 : 3 `
149.

If, 25 C of charge is determined to pass through a wire of any cross section in 50s, what is the measure of current?

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SOLUTION :`I= Q"/"t=(25C)"/"(50S)=0.5C"/"s=0.5A`
150.

The unit of work is Joule. The other physical quantity that has same unit is

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Power
Velocity
energy
Force

Answer :C