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InterviewSolution
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.
| 5051. |
A runner completes one round of a circular track of diameter 400 m in 40 s. What will be the distance covered and the displacement respectively at the end of 2 min 20 s? |
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Answer» A runner completes one round of a circular track of diameter 400 m in 40 s. What will be the distance covered and the displacement respectively at the end of 2 min 20 s? |
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| 5052. |
On which of the following factors does the kinetic energy of a body depend? |
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Answer» On which of the following factors does the kinetic energy of a body depend? |
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| 5053. |
Question 2A car is accelerated on a leveled road and attains a velocity 4 times of its initial velocity. In this process the potential energy of the car(a) does not change(b) becomes twice to that of initial(c) becomes 4 times that of initial(d) becomes 16 times that of initial |
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Answer» Question 2 A car is accelerated on a leveled road and attains a velocity 4 times of its initial velocity. In this process the potential energy of the car (a) does not change (b) becomes twice to that of initial (c) becomes 4 times that of initial (d) becomes 16 times that of initial |
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| 5054. |
Question 22(ii)An electron moving with a velocity of 5×104ms−1 enters into a uniform electric field and acquires a uniform acceleration of 104ms−2 n the direction of its initial motion.(ii) How much distance the electron would cover in this time? |
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Answer» Question 22(ii) An electron moving with a velocity of 5×104ms−1 enters into a uniform electric field and acquires a uniform acceleration of 104ms−2 n the direction of its initial motion. (ii) How much distance the electron would cover in this time? |
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| 5055. |
What is vactor law and magnitude |
| Answer» What is vactor law and magnitude | |
| 5056. |
निम्नलिखित प्रश्न के उत्तर (25-30 शब्दों में) लिखिए -लेखक ने राजकपूर को एशिया का सबसे बड़ा शोमैन कहा है। शोमैन से आप क्या समझते हैं? |
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Answer» निम्नलिखित प्रश्न के उत्तर (25-30 शब्दों में) लिखिए - लेखक ने राजकपूर को एशिया का सबसे बड़ा शोमैन कहा है। शोमैन से आप क्या समझते हैं? |
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| 5057. |
8. From the following molar conductivities at infinite dilution, ∧^° for Al_2(SO_4)_3=858 S cm^2 mol^{-1} ∧^° for NH_4OH = 238.3 S cm^2 mol^{-1} ∧^° for (NH_4)_2SO_4 = 238.4 S cm^2 mol^{-1} Calculate ∧^° for Al(OH)_3 |
| Answer» 8. From the following molar conductivities at infinite dilution, ∧^° for Al_2(SO_4)_3=858 S cm^2 mol^{-1} ∧^° for NH_4OH = 238.3 S cm^2 mol^{-1} ∧^° for (NH_4)_2SO_4 = 238.4 S cm^2 mol^{-1} Calculate ∧^° for Al(OH)_3 | |
| 5058. |
Find the resistivity of the given resistor of resistance 25 Ω? |
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Answer» Find the resistivity of the given resistor of resistance 25 Ω? |
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| 5059. |
If a lens is made up of two different materials then there are two different refractive indices. But how does it has 2 different foci? |
| Answer» If a lens is made up of two different materials then there are two different refractive indices. But how does it has 2 different foci? | |
| 5060. |
acceleration due to gravity at the surface of the planet is equal to that of surface of earth and density is 1.5times that of earth. if radius of earth is R, radius of planet is1. 3/2R2. 2/3R3. 9/4R 4. 4/9R |
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Answer» acceleration due to gravity at the surface of the planet is equal to that of surface of earth and density is 1.5times that of earth. if radius of earth is R, radius of planet is 1. 3/2R 2. 2/3R 3. 9/4R 4. 4/9R |
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| 5061. |
dimensional formula for mass, if kinetic energy, universal gravitational constant and angular momentum are taken as base quantities is |
| Answer» dimensional formula for mass, if kinetic energy, universal gravitational constant and angular momentum are taken as base quantities is | |
| 5062. |
How many moons does Mars have? |
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Answer» How many moons does Mars have? |
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| 5063. |
What is rate of change of speed? |
| Answer» What is rate of change of speed? | |
| 5064. |
Derivation of the 1st equation of motion |
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Answer» Derivation of the 1st equation of motion |
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| 5065. |
Use the following information to answer the next question. The force of attraction between two bodies having masses 4.5 × 106 kg and m is 1.24 × 10−4 N. These bodies are separated by a distance of 2.5 × 103 m. What is the value of m? |
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Answer» Use the following information to answer the next question. The force of attraction between two bodies having masses 4.5 × 106 kg and m is 1.24 × 10−4 N. These bodies are separated by a distance of 2.5 × 103 m. What is the value of m? |
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| 5066. |
09(e)4-)\}The position-time graph for two particles A and Bare straight lines inclined at angles of 45 and 60"with time axis as shown in figure. The magnituderelative velocity of A w.r.t B is(w)x45(s)RO(1) (2-1) m/s(2) 3 m/s(3) (3-1) m/s(4) 1 m/s |
| Answer» 09(e)4-)\}The position-time graph for two particles A and Bare straight lines inclined at angles of 45 and 60"with time axis as shown in figure. The magnituderelative velocity of A w.r.t B is(w)x45(s)RO(1) (2-1) m/s(2) 3 m/s(3) (3-1) m/s(4) 1 m/s | |
| 5067. |
Two satelite of mass m and 2m are revolving in two circular orbit of radius r and 2r around an imaginary planet on the surface of which gravitational force is inversely proportional to distance from its cwmtre. The ratio of orbital speed of satelite is 1. 1:12. 1:23. 2:14. 1:√2 |
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Answer» Two satelite of mass m and 2m are revolving in two circular orbit of radius r and 2r around an imaginary planet on the surface of which gravitational force is inversely proportional to distance from its cwmtre. The ratio of orbital speed of satelite is 1. 1:1 2. 1:2 3. 2:1 4. 1:√2 |
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| 5068. |
A stone dropped from a window reaches the ground in 0.5 s (given g=10 m/s^2) (i) Calculate the speed just before it hits the ground. (ii) What is the average speed at t=0.5 s? (iii) Calculate the height of window from the ground. |
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Answer» A stone dropped from a window reaches the ground in 0.5 s (given g=10 m/s^2) (i) Calculate the speed just before it hits the ground. (ii) What is the average speed at t=0.5 s? (iii) Calculate the height of window from the ground. |
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| 5069. |
When traffic light turns green, a car moves with a cons†an t acceleration 2 m/s^2. At same time a truck moving with a cons†an t speed of 10 m/s, overtakes and passes the car. (1) How far beyond the starting point will car overtake the truck? (2) How fast will car be travelling at that ins†an t? |
| Answer» When traffic light turns green, a car moves with a cons†an t acceleration 2 m/s^2. At same time a truck moving with a cons†an t speed of 10 m/s, overtakes and passes the car. (1) How far beyond the starting point will car overtake the truck? (2) How fast will car be travelling at that ins†an t? | |
| 5070. |
What is the equivalent resistance between A and B? |
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Answer» What is the equivalent resistance between A and B? |
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| 5071. |
An impulse →J changes the velocity of particle from →v1 to →v2. Change in kinetic energy of the particle isएक आवेग →J कण के वेग को →v1 से →v2 तक परिवर्तित करता है। कण की गतिज ऊर्जा में परिवर्तन है |
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Answer» An impulse →J changes the velocity of particle from →v1 to →v2. Change in kinetic energy of the particle is |
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| 5072. |
Figure given below is the distance-time graph of the straight line motion of an object.(a) What will be the position of the object at 20 s?(b) What will be the distance travelled by the object in 12 s?(c) What is the average speed of the object? |
Answer» Figure given below is the distance-time graph of the straight line motion of an object. (a) What will be the position of the object at 20 s? (b) What will be the distance travelled by the object in 12 s? (c) What is the average speed of the object? |
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| 5073. |
The electric potential at a point (x, y, z) is given by V = – xy2 – x2z3 + 4. The electric field at that point is |
| Answer» The electric potential at a point (x, y, z) is given by V = – xy2 – x2z3 + 4. The electric field at that point is | |
| 5074. |
A particle is projected vertically upward with velocity v from the top of tower of height H. Its velocity at a depth h below the point of projection is thrice of its velocity at a height h above the point of projection . Find the time at which it hitthe ground. Also find the velocity with which the particle hit the ground.Find the maximum hight achieved by the particle above the top of the tower in term of h. |
| Answer» A particle is projected vertically upward with velocity v from the top of tower of height H. Its velocity at a depth h below the point of projection is thrice of its velocity at a height h above the point of projection . Find the time at which it hitthe ground. Also find the velocity with which the particle hit the ground.Find the maximum hight achieved by the particle above the top of the tower in term of h. | |
| 5075. |
Definition of direction |
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Answer» Definition of direction |
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| 5076. |
If equal volumes of water and alcohol are heated to the same rise of temperature, then |
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Answer» If equal volumes of water and alcohol are heated to the same rise of temperature, then |
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| 5077. |
A cord is used to lower vertically a block of mass M by a dis†an ce d at a cons†an t downward acceleration of g/4.Then the work done by the cord on the block is (a)Mgd/4 (b)3Mgd/4 (c)-3Mgd/4 (d)Mgd? |
| Answer» A cord is used to lower vertically a block of mass M by a dis†an ce d at a cons†an t downward acceleration of g/4.Then the work done by the cord on the block is (a)Mgd/4 (b)3Mgd/4 (c)-3Mgd/4 (d)Mgd? | |
| 5078. |
two tuning forks have freq. 540 and 524Hz. On sounding them together the interval between two successive maximum intensity? |
| Answer» two tuning forks have freq. 540 and 524Hz. On sounding them together the interval between two successive maximum intensity? | |
| 5079. |
2 point light source are 24cm apart, where should a convex lens of focal length 9cm be put in b/w them 1 source so that image of both the source are formed at same place? 6cm,9cm,12cm,15cm |
| Answer» 2 point light source are 24cm apart, where should a convex lens of focal length 9cm be put in b/w them 1 source so that image of both the source are formed at same place? 6cm,9cm,12cm,15cm | |
| 5080. |
From a rifle of mass 4kg,a bullet of mass 50g is fired with a initial velocity of 35 m/s.Calculate the initial recoil velocity of the rifle. |
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Answer» From a rifle of mass 4kg,a bullet of mass 50g is fired with a initial velocity of 35 m/s.Calculate the initial recoil velocity of the rifle. |
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| 5081. |
What is thermal power plant? Explain the principle of working of thermal power plant. |
| Answer» What is thermal power plant? Explain the principle of working of thermal power plant. | |
| 5082. |
Light waves are travelling in vaccum, along the y-axis. Which of the following, may represent a plane wavefront? |
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Answer» Light waves are travelling in vaccum, along the y-axis. Which of the following, may represent a plane wavefront? |
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| 5083. |
A man is moving with velocity 6 m/s towards east experiences that is falling vertically down if he increases his speed to 12 m/s. Now if he experiences the rain at 37 degree with vertical.Then find real velocity of rain? |
| Answer» A man is moving with velocity 6 m/s towards east experiences that is falling vertically down if he increases his speed to 12 m/s. Now if he experiences the rain at 37 degree with vertical.Then find real velocity of rain? | |
| 5084. |
47. Total K.E. of sphere of mass M rolling with velocity v is a) 7/10 Mv² b) 5/6 Mv² c) 7/5 Mv² d) 10/7 Mv² Click here to view details: |
| Answer» 47. Total K.E. of sphere of mass M rolling with velocity v is a) 7/10 Mv² b) 5/6 Mv² c) 7/5 Mv² d) 10/7 Mv² Click here to view details: | |
| 5085. |
11. A AIR bubble rise from bottom of a lake to surface. If its radius increases by 200%and atmospheric pressure is equal to water column of height H, then depth if lake is? |
| Answer» 11. A AIR bubble rise from bottom of a lake to surface. If its radius increases by 200%and atmospheric pressure is equal to water column of height H, then depth if lake is? | |
| 5086. |
39.An object floats in water(at 4^° C) with 20% of its volume outside the water surface. What percentage of volume of the same body will be outside the liquid surface whose relative density in5? |
| Answer» 39.An object floats in water(at 4^° C) with 20% of its volume outside the water surface. What percentage of volume of the same body will be outside the liquid surface whose relative density in5? | |
| 5087. |
A force of magnitude 25 N acting on a body of mass 2 kg increases its kinetic energy from 100 J to 200 J. The displacement of the body during this interval is ___. |
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Answer» A force of magnitude 25 N acting on a body of mass 2 kg increases its kinetic energy from 100 J to 200 J. The displacement of the body during this interval is ___. |
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| 5088. |
A stone is dropped into a deep well and is heard to hit the water 2 s after being dropped. Determine the depth of the well. Assume no time delay in stone hitting the water and sound being heard. Take a=10ms2. |
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Answer» A stone is dropped into a deep well and is heard to hit the water 2 s after being dropped. Determine the depth of the well. Assume no time delay in stone hitting the water and sound being heard. Take a=10ms2. |
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| 5089. |
Intensity of Electric field of a dipole at any point in infinity |
| Answer» Intensity of Electric field of a dipole at any point in infinity | |
| 5090. |
A force of 10 N is applied on an object of mass 1 kg for 2s, which was initially at rest. What is the work done on the object by the force ? |
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Answer» A force of 10 N is applied on an object of mass 1 kg for 2s, which was initially at rest. What is the work done on the object by the force ? |
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| 5091. |
Water vapours have more energy than water at same twtemperatu . Why?? |
| Answer» Water vapours have more energy than water at same twtemperatu . Why?? | |
| 5092. |
41. In damped oscillations, the amplitude of oscillations is reduced to half of its initial value of5 cm at the end of 25 oscillations. What will be its amplitude when the oscillator completes50 oscillations? |
| Answer» 41. In damped oscillations, the amplitude of oscillations is reduced to half of its initial value of5 cm at the end of 25 oscillations. What will be its amplitude when the oscillator completes50 oscillations? | |
| 5093. |
A cricket ball of mass 200 g moving with a speed of 40 ms−1 is brought to rest by a player in 0.04s. Calculate the following : (i) change in momentum of the ball, (ii) average force applied by the player. [2 MARKS] |
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Answer» A cricket ball of mass 200 g moving with a speed of 40 ms−1 is brought to rest by a player in 0.04s. Calculate the following : |
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| 5094. |
An object of mass 20 kg is moving with a velocity of 10 m/s. Its momentum will be: |
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Answer» An object of mass 20 kg is moving with a velocity of 10 m/s. Its momentum will be: |
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| 5095. |
Study the entries in the following table and rewrite them putting the connected items in a single row. I II III Mass m/s2 Zero at the centre Weight kg Measure of inertia Acceleration due to gravity Nm2/kg2 Same in the entire universe Gravitational constant N Depends on height |
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Answer» Study the entries in the following table and rewrite them putting the connected items in a single row.
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| 5096. |
A stone is dropped into lake from a tower 500m high. The sound of the splash will be heard by a man on the tiwer after? ( Speed of sound=340m/s) |
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Answer» A stone is dropped into lake from a tower 500m high. The sound of the splash will be heard by a man on the tiwer after? ( Speed of sound=340m/s) |
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| 5097. |
The energy of 4900 joule was expanded in lifting a 50 kilogram of mass what is the height? |
| Answer» The energy of 4900 joule was expanded in lifting a 50 kilogram of mass what is the height? | |
| 5098. |
94. A circular current carrying loop is an equivalent to a magnetic dipole. Now a point on the axis of the loop is 1) a broad side on position 2) an end on position 3) both 1 and 2 4) neither 1 amd 2 |
| Answer» 94. A circular current carrying loop is an equivalent to a magnetic dipole. Now a point on the axis of the loop is 1) a broad side on position 2) an end on position 3) both 1 and 2 4) neither 1 amd 2 | |
| 5099. |
Explain the term ‘gravity’. |
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Answer» Explain the term ‘gravity’. |
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| 5100. |
A simple pendulum is placed on the surface of Moon. While oscillating it rises to a maximum height of 5 cm from its mean position. If the mass of bob is 500 gram and acceleration due to gravity on Moon is g=2 ms−2. Find the below: i) The total energy of simple pendulum at any instant while oscillating. ii) The velocity of bob at its mean position. (Assume its mean position is at ground level) |
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Answer» A simple pendulum is placed on the surface of Moon. While oscillating it rises to a maximum height of 5 cm from its mean position. If the mass of bob is 500 gram and acceleration due to gravity on Moon is g=2 ms−2. Find the below: |
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