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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. |
Calculate the force required to impart a car a velocity of `30m//s`, in 10 seconds. The mass of the car is `1500kg` |
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Answer» Here, Mass, m = 1500 kg Let us calculate the value of acceleration by using the first equation of motion. Now, Initial velocity, u = 0 (Car starts from rest) Final velocity, v = 30 m/s And Time taken, t = 10s Now, putting these values in the equation: v=u+at We get, `30 = 0+axx 10` 10a=30 `a=30/10m//s^(2)` or Acceleration,a=3m/`s^(2)` Now, putting m=1500 kg and a=3 m/`s^(2)` in equation: `F=mxxa` We get, F= 1500 `xx` 3 N = 4500 N Thus, the force required in this case is of 4500 newtons. |
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| 2. |
A force of `5N` gives a mass `m_(1)`, an acceleration of `8 m//s^(2)`, and a mass `m_(2)`, an acceleration of `24 m//s^(2)`. What acceleration would it give if both the masses are tied together? |
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Answer» (i) In the first case: Force, F=5 N Mass,m=`m_(1)` (To be calculated) And, Acceleration, a = 8 m/`s^(2)` Now, F=`mxxa` So, `5=m_(1)xx8` And, `m_(1)=5/8kg` `m_(1)=0.625kg` Thus, the mass `m_(1)` is 0.625 kg. (ii) In the second case: Force, F = 5 N Mass, m=`m_(2)` (To be calculated) And, Acceleration a=24 m/`s^(2)` Now, F=m`xx`a So, `5=m_(2)xx24` And, `m_(2)=5/24kg` `m_(2)=0.208` kg Thus, the mass m2 is 0.208 kg. (iii) In the third case: Force=5N Total mass, m = `m_(1)+m_(2)`. =0.625+0.208 =0.833 kg And, Acceleration, a = ? (To be calculated) Now putting these values in the relation : `F=mxxa` We get : `5=0.833xxa` `a=5/0.833` `a=6 m//s^(2)` Thus, if both the masses are tied together, then the acceleration would be 6 m/`s^(2)` |
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| 3. |
Which would require a greater force - accelerating a 10 g mass at 5 m/ `s^(2)` or 20 g mass at 2 m/`s^(2)`? |
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Answer» (i) In first case: Force, F=m`xx`a =`10/1000kgxx5m//s^(2)` =0.05N (ii) In second case : Force,F=`20/1000kgxx2m//s^(2)` =0.04 N Thus, a greater force of 0.05 N is required for accelerating a 10 g mass. |
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| 4. |
A hammer of mass `500g`, moving at `50m//s`, strikes a nail. The nail stops the hammer in a very short time of `0.01 s`. What is the force of the nail on the hammer? |
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Answer» Here, Mass,m=500 g `=500/1000Kg` =0.5 kg We will now calculate the acceleration. Now, Initial velocity,`u=50 m1s^(-1)` Final velocity,v = 0 (The hammer stops) Acceleration, a= ? (To be calculated) And, Time,t=0.01 s Now, v=u+at So, `0=50+axx0.01` 0.01a=-50 `a=-50/0.01` `a=-5000ms^(-2)` We know that : Force,F=m`xx`a So, `F=0.5xx(-5000)N` F=-2500 N Thus, the force of the nail on the hammer is 2500 newtons. The negative sign indicates that this force is opposing motion. |
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| 5. |
A truck starts from rest and rolls down a hill with a constant acceleration. It travels a distance of `400m` in `20 s`. Find its acceleration. Find the force acting on it if its mass is `7` metric tonnes (Hint.1 metric tonne=1000kg) |
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Answer» (a) Calculation of acceleration Here, Initial velocity, u=0 Distance travelled, s=400 m Time, t= 20 s And Acceleration, a =? (To be calculated) Now, `s=ut+1/2at^(2)` so, `400=0xx20+1/2xxaxx(20)^(2)` `400=0+1/2xxaxx400` `400=axx200` `a=400/200` `a=2m//s^(2)` Thus the acceleration of the truck is `2 m//s^(2)` (b) Calculation of force Force,F=m`xx`a F=`7xx1000xx2` F=14000N Thus, the force acting on the truck is of 14000 newtons. |
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| 6. |
A stone of 1 kg is thrown with a velocity of `20ms^(-1)` across the frozen surface of a lake and comes to rest after travelling a distance of 50 m. What is the force of friction between the stone and the ice ? |
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Answer» Here, Initial velocity, u=`20ms^(-1)` final velocity,v=0 (The stone stops) Acceleration,a = ? (To be calculated) And, Distance travelled, s=50 m Now, `v^(2)=u^(2)+2as` `(0)^(2)=(20)^(2)+2xxaxx50` 0-400+100 a 100 a=-400 `a=-400/100` `a=-4ms^(-2)` Now, Force,F=`mxxa` `F=1xx(-4)N` F=-4N Thus, the force of friction between the stone and the ice is 4 newtons. The negative sign shows that this force opposes the motion of stone. |
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| 7. |
Two objects of masses `100g` and `200g` are moving along the same line in the same direction with velocities of `2m//s` and `1m//s`, respectively. They collide and after the collison, the first object moves at a velocity of `1.67 m//s` in the same direction. Determine the velocity of the second object. |
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Answer» In order to solve this problem, we will first calculate total momentum of both the objects before and after the collision. Momentum of first object (before collision)=Mass of first object`xx`Velocity of first object `=100/1000kgxx2ms^(-1)` `0.1kgxx2ms^(-1)` `=0.2 kg ms^(-1)` Momentum of second object (before collision) = Mass of second object`xx` Velocity of second object `=200/1000kgxx1ms^(-1)` `=0.2kgms^(-1)` Total momentum = 0.2 + 0.2 (before collision)=-04 kg `m s^(-1)` (b) After collision, the velocity of first object of mass 100 g becomes 1.67 m `s^(-1)`. So, Momentum of first object (after collision)=`100/1000kgxx1.67ms^(-1)` `=0.1kgxx1.67ms^(-1)` `=0.167kgms^(-1)` After collision, suppose the velocity of second object of mass 200 g becomes v`ms^(-1)`. So, Momentum of second object (after collision)=`200/1000kgxxvms^(-1)` `=0.2kgxxvms^(-1)` `=0.2vkgms^(-1)` Total momentum (after collision)=0.167+0.2 v Now, according to the law of conservation of momentum : Total momentum before collision=Total collision after collision That is, 0.4=0.167+0.2v 0.2v=0.4-0.167 0.2v=0.233 `v=0.233/0.2` `v=1.165ms^(-1)` Thus, the velocity of second object is 1.165 metres per second. |
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| 8. |
In the following example,try to identify the number of times the velocity of the ball changes: "A football player kicks a football to another player of his team who kicks the football towards the goal. The goalkeeper of the opposite team collects the football and kicks it towards a player of his own team"? Also identify the agent supplying the force in each case. |
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Answer» Correct Answer - (i) The velocity of football changes for the 1st time when a football player kicks the football to another player of his team. The agent supplying the force in this case is the kick applied by the player (i) The velocity of football changes for the 2nd time when another player of the same team kicks the football towards the goal. Here the force is supplied by the kick of another player. (iii) The velocity of football changes for the 3rd time when the goalkeeper of opposite tean collects the football (or stops the football). In this case the force is applied by the hands of the goalkeeper (iv) The velocity of football changes for the 4th time when the goalkeeper kicks the stationary football towards a player of his own team. Here the force is supplied by the kick of goalkeeper. Thus, the velocity of football changes four times. |
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| 9. |
A water tanker filled up to two-thirds of its tank with water is running with a uniform speed. When the brakes are suddenly applied, the water in its tank would:A. move backwardB. move forwardC. rise upwardsD. remain unaffected |
| Answer» Correct Answer - b | |
| 10. |
A passenger in a moving train tosses a coin which falls behind him. It means that motion of the train isA. acceleratedB. uniformC. retardedD. along circular track |
| Answer» Correct Answer - a | |
| 11. |
When a rubber balloon held between the hands is pressed, its shape changes. This happens because :A. balanced forces act on the balloonB. unbalanced forces act on the balloonC. frictional forces act on the balloonD. gravitational forces act on the balloon |
| Answer» Correct Answer - a | |
| 12. |
The SI unit of momentum isA. kg-m/sB. gm-cm/sC. gm-cm/minD. kg-cm/s |
| Answer» Correct Answer - A | |
| 13. |
The inertia of a moving object depends on:A. momentum of the objectB. speed of the objectC. mass of the objectD. shape of the object |
| Answer» Correct Answer - c | |
| 14. |
A bullet of mass 10 g is fired from a gun of mass 6 kg with a velocity of 300 m/s. Calculate the recoil velocity of the gun. |
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Answer» Here, Mass of bullet=10 g =`10/100` kg =0.01 kg Velocity of bullet=300 m/s Mass of gun=6kg And, Recoil velocity of gun=? (To be calculated) Now, putting these values in the relation : Mass of bullet `xx`Velocity of bullet`xx`=mass of gun `xx`Recoil velocity of gun We get : `0.01xx300=6xx` Recoil velocity of gun So, Recoil velocity of gun=`(0.01xx300)/6` =0.5 m/s Note. The above prblem can also be solved by calculating the momentum of bullet and the gun separately as follows : Momentum of bullet=Mass of bullet`xx`Velocity of bullet `=0.01xx300` = 3 kg. m/s Now, suppose the recoil velocity of gun is v m/s. So, Momentum of gun=Mass`xx`Recoil velocity of gun `=6xx`v kg m/s According to the law of conservation of momentum : So, `3=6xxv` And, `v=3/6m//s` Recoil velocity of gun,v=0.5 m/s |
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| 15. |
A bullet of mass 10 g moving with a velocity of 400 m/s gets embedded in a freely suspended wooden block of mass 900 g. What is the velocity acquired by the block ? |
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Answer» Here, Mass of the bullet,`m_(1)=10g` `=10/1000`kg =0.01 kg And, Velocity of the bullet, `v_(1)`=400 m/s So, Momentum of bullet=`m_(1)xxv_(1)` `=0.01xx400kg.m//s`…….(1) Now, this bullet of mass 10 g gets embedded into a wooden block of mass 900 g . So ,the mass of wooden blockk alongwith the embedded bullet will become 900+10=910 g. Thus, Mass of wooden block+Bullet, `m_(2)=900+10` =910 g `=910/1000` kg =0.91 kg And, Velocity of wooden block+bullet,`v_(2)=?` (To be calculated) So, Momentum of wooden block+bullet=`m_(2)xxv_(2)` `=0.91xxv_(2)`kg.m/s...(2) Now, according to the law of conservation of momentum, the two momenta as given by equations (1) and (2) should be equal. So, `m_(1)xxv_(1)=m_(2)xxv_(2)` or 0.01`xx400=0.91xxv_(2)` And, `v_(2)=(0.01xx400)/0.91` =4.4 m/s Thus, the velocity acquired by the wooden block (having the bullet embedded in it) is 4.4 metres per second. |
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