Problems Based On Gravitation and Gravity: Formula Used: (i) \(F = \frac{{GMm}}{{{R^2}}}\) (ii) Acceleration due to gravity (g) = \(\frac{{...
Problems Based On Gravitation and Gravity:
Formula Used:
(i) \(F = \frac{{GMm}}{{{R^2}}}\)
(ii) Acceleration due to gravity (g) = \(\frac{{GM}}{{{R^2}}}\)
Where M is the mass of the Earth and R is the radius of the earth.
Where M is the mass of the Earth and R is the radius of the earth.
(1) Consider a heavenly body whose mass is twice that of the earth and whose radius is thrice that of the earth. What will be the weight of a book on this heavenly body, if its weight on the earth is 900 N.
Ans: 200 N
(2) Find the gravitational force between two protons kept at a separation of 1 femtometre. (1 femtometre = \({10^{ - 15}}m\)). The mass of a proton is \(1.67 \times {10^{ - 27}}kg\).
Ans: \(1.86 \times {10^{ - 34}}N\)
(3) Find the gravitational force between the sun and the earth. The mass of the sun is \(2 \times {10^{30}}kg\) and the mass of the earth is \(6 \times {10^{24}}kg\). The distance between the sun and the earth is \(1.5 \times {10^{11}}m\)
Ans: \(3.56 \times {10^{22}}N\)
(4) Consider a heavenly body whose mass is \(3 \times {10^{24}}kg\) (half that of the earth) and radius is 3200 km (half that of the earth). What is the acceleration due to gravity at the surface of this heavenly body?
Ans: \(19.6m.{\sec ^{ - 2}}\)
(5) Two bodies 'A' and 'B' of masses 'm' and '2m' respectively are kept a distance 'd' apart. Where should a small particle be placed so that the net gravitational force on it due to the bodies 'A' and 'B' is zero?
Ans: \(x = \frac{d}{{1 + \sqrt 2 }}\)
(6) Two bodies of masses 1 kg and 2 kg respectively are placed at a separation of 1 m. Find the acceleration of the bodies assuming that only gravitational force act.
Ans: \(1.33 \times {10^{ - 10}}m.{\sec ^{ - 2}}\), \(6.67 \times {10^{ - 11}}m.{\sec ^{ - 2}}\)
(7) Communication satellites move in orbit of radius 44,400 km around the earth. Find the acceleration of such a satellite, assuming that the only force acting on it is that due to the earth. Mass of the earth \(6 \times {10^{24}}kg\).
Ans: \(0.2m.{\sec ^{ - 2}}\)
(8) Suppose an astronaut lands on the moon and drops an object from a height of 7.35 m from the surface. How much time will it take to reach the moon's surface?
Ans: \(3\sec \)
(9) A body of mass \(1kg\) is placed at a distance of \(2m\) from another body of mass \(10kg\). At what distance from the body of \(1kg\) , another body of mass \(5kg\) be placed so that the net force of gravitation acting on the body of mass \(1kg\) is zero?
Ans: \(\sqrt 2 m\)
(10) What is the magnitude of the gravitational force between the earth and a \(1kg\) object on its surface? Mass of the earth \(6 \times {10^{24}}kg\) and radius of the earth \(6.4 \times {10^6}m\)
Ans: \(9.8N\)
(11) Calculate the force of gravitation between the earth and the sun, given that the mass of the earth \(6 \times {10^{24}}kg\) and of the sun \(2 \times {10^{30}}kg\). The average distance between the two is \(1.5 \times {10^{11}}m\).
Ans: \(3.56 \times {10^{22}}N\)
(12) Calculate the force of gravitation between the earth and the sun, given that the mass of the earth \(6 \times {10^{24}}kg\) and the sun \(2 \times {10^{30}}kg\). The average distance between the two is \(1.5 \times {10^{11}}m\)
Ans:
(13) Derive the unit of force using the second law of motion. A force of \(5\sec \) produces an acceleration of \(8m.{\sec ^{ - 2}}\) on a mass \({m_1}\) and an acceleration of \(24m.{\sec ^{ - 2}}\) on a mass \({m_2}\). What acceleration would the same force provide if both the masses are tied together?
Ans:
(15) Calculate the gravitational force between a 10 kg ball and a 20 kg ball placed at a separation of 5 m.
Ans: \(5.34 \times {10^{ - 10}}N\)
(16) Three balls A, B and C are kept in a straight line. The separation between 'A' and 'C' is 1 m and 'B' is placed at the midpoint between them. The masses of A, B and C are 100 gm, 200 gm and 300 gm respectively. Find the net gravitational force on (i) A (ii) B and (iii) C.
Ans: \(7.34 \times {10^{ - 12}}N\), \(1.07 \times {10^{ - 11}}N\), \(1.80 \times {10^{ - 11}}N\)
(17) A particle of mass \({m_1}\) is kept at \(x = 0\) and another of mass \({m_2}\) at \(x = d\). When a third particle is kept \(x = \frac{d}{4}\) , it experience no net net gravitational force due to two particle. Find \(\frac{{{m_2}}}{{{m_1}}}\)
Ans: 9
(18) The acceleration due to gravity near the earth's surface is \(9.8m.{\sec ^{ - 2}}\) and the earth 's radius is 6400 km. From this data calculate the mass of the earth. Use any universal constant if required.
Ans: \(6.02 \times {10^{^{24}}}kg\)
(19) Two particle of mass 200 gm each are placed at a separation of 10 cm. Assume that the only force acting on them are due to their gravitational attraction of each when they are allowed to move.
Ans: \(1.33 \times {10^{ - 9}}m.{\sec ^{ - 2}}\), \(1.33 \times {10^{ - 9}}m.{\sec ^{ - 2}}\)
(20) A particle weighs 120 N on the surface of the earth. At what height above the earth's surface will its weight weight be 30 N? Radius of the earth = 6400 Km
Ans: 6400 Km
(21) Suppose the earth shrinks such that its radius decreases to half the present value. What will be the acceleration due to gravity on the surface of the earth.
Ans: \(39.2m.{\sec ^{ - 2}}\)
(22) A body weigh 120 N. on the surface of the earth. Find its approximate weight on the moon.
Ans: 20 N
(23) Calculate the value of the acceleration due to gravity due to gravity at a place 3200 Km above the surface of the earth.
Ans: \(4.36m.{\sec ^{ - 2}}\)
Problems Based on Variation of Acceleration due to height, depth from the surface of the Earth:
Formula Used:
(i) \({g^'} = \frac{{g{R^2}}}{{{{\left( {R + h} \right)}^2}}}\)
(ii) \({g^'} = g\left( {1 - \frac{d}{R}} \right)\)
(24) Weight of an object is \(294N\) on the surface of the earth. What is its weight at a height of \(200Km\) from the surface of the earth. Radius of the earth =\(6400km\)
Ans: \(276.45N\)
(25) Communication satellites move in orbits of radius \(44,400km\) around the earth. Find the acceleration of such a satellite, assuming that the only force acting on it is that due to the earth. Mass of earth \(6 \times {10^{24}}kg\)
Ans: \(0.2m.{s^{ - 2}}\)
(26) How much below the surface of earth does the acceleration due to gravity (i) reduce to \(36\% \) (ii) reduce by \(36\% \) of its value on the surface of the earth, Radius of earth \(6400km\)
Ans: \(4096km\), \(2304km\)
(27) Find the value of acceleration due to gravity at a height of \(12,800km\) from the surface of the earth. Earth's radius \(6400km\)
Ans: \(1.09m.{s^{ - 2}}\)
(29) An object weighs \(294N\) on the earth. (i) What would be its mass on the moon? (ii) What is the acceleration due to gravity on the moon?
Ans:
(31) Calculate the acceleration due to gravity on the surface of the moon (mass of the moon \(7.4 \times {10^{22}}kg\) and radius \(1.74 \times {10^6}m\) )
Ans:
(32) The weight of a body on the surface of earth is \(392N\). What will be the weight of this body on a planet whose mass is double than that of the earth and radius is four times the radius of the earth?
Ans:
(33) Find the value of the acceleration due to gravity at a height of 12,800 km from the surface of the earth. Earth's radius = 6400 km.
Ans: \(1.09m.{\sec ^{ - 2}}\)
(34) Find the weight of an object at a height 6400 km above the earth's surface. The weight of the object at the surface of the earth is 20 N and radius of the earth is 6400 km.
Ans: 5 N
(35) A particle is taken to a height of \(2{R_e}\) above the earth's surface, where \({R_e}\) is the radius of the earth. If it is dropped from this height, what would be its acceleration?
Ans: \(1.1m.{\sec ^{ - 2}}\)
(36) The acceleration due to gravity at a place is \(0.2m{\sec ^{ - 2}}\). Find its height above the earth's surface.
Ans: \(6{R_e} = 38,400Km\)
(37) As one moves to a place 3200 Km above the earth's surface, the acceleration due to gravity reduce to \(\frac{4}{9}th\) of its value at the earth's surface. Calculate the radius of the earth from this data.
Ans: 6400 Km
Problems Based on Acceleration due to gravity:
Formula Used:
| SL NO | Kinematics with acceleration ‘a’ | Kinematics with acceleration ‘g’ |
|---|---|---|
| 1 | \(v = u + at\) | \(v = u + gt\) |
| 2 | \(s = ut + \frac{1}{2}a{t^2}\) | \(s = ut + \frac{1}{2}g{t^2}\) |
| 3 | \({v^2} = {u^2} + 2as\) | \({v^2} = {u^2} + 2gs\) |
| 4 | \({s_n} = u + \frac{1}{2}a\left( {2n - 1} \right)\) | \({s_n} = u + \frac{1}{2}g\left( {2n - 1} \right)\) |
(38) A ball is dropped from the edge of a roof. It takes \(0.1\sec \) to cross a window of height \(2.0m\). Find the height of the roof above the top of the window.
Ans: \(19.6m\)
(39) A stone is dropped from the edge of the roof. (i) How long does it take to fall \(4.9m\)? (ii) How fast does it move at the end of the fall? (iii) How fast does it move at the end of \(7.9m\)? (iv) What is its acceleration after \(1\sec \) and after \(2\sec \)?
Ans: \(1\sec \), \(9.8m/s\), \(12.4m/\sec \), \(9.8m.{s^{ - 2}}\), \(9.8m.{s^{ - 2}}\)
(40) A ball is thrown vertically upwards with a velocity of \(49m/s\). Calculate (i) the maximum height to which it rises (ii) the total time it take to return to the surface of the earth.
Ans: \(122.5m\), \(10\sec \)
(41)A stone is released from the top of a tower of height \(19.6m\). Calculate its final velocity just before touching the ground.
Ans: \(19.6m/s\)
(42) A stone is thrown vertically upward with an initial velocity of \(40m/s\). Taking \(g = 10m.{s^{ - 2}}\), find the maximum height reached by the stone. What is the net displacement and the total distance covered by the stone.
Ans: \(80m\), \(160m\)
(43) A stone is allowed to fall from the top of a tower \(100m\) high and the same time another stone is projected vertically upwards from the ground with a velocity of \(25m.{s^{ - 2}}\). Calculate when and where the two stone will meet.
Ans: \(21.6m\)
(44) A ball thrown up vertically returns to the after \(6\sec \). Find (i) the velocity with which it was thrown up (ii) the maximum height it reaches, and (iii) its position after \(4\sec \).
Ans: \(29.4m\) \(44.1m\), \(4.9m\)
(45) A boy drops a ball from the top of a tower of height \(19.6m\). Calculate the velocity of the ball just before it touches the ground?
Ans:
(46) A ball is thrown vertically upwards with a velocity of \(49m/\sec \). Calculate (i) the maximum height to which it rises. (ii) the total time it takes to return to the surface of the earth.
Ans:
(47) What is the force of gravity acting on a point mass of \(1kg\) released from from a height of \(1m\) from the surface of the earth? The radius of the earth \(6.4 \times {10^6}m\), mass of earth \(6 \times {10^{24}}kg\)
Ans:
(48) A stone is released from the top of a tower of height \(19.6m\). Calculate its final velocity just before touching the ground.
Ans:
(49) When a ball is thrown vertically upwards, it goes through a distance of \(19.6m\). Find the initial velocity of the ball and the time taken by it to rise to the highest point. (Acceleration due to gravity \(g = 9.8m.{\sec ^{ - 2}}\) )
Ans:
(50) A ball thrown up is caught back by the thrower after \(6\sec \). Calculate (i) the velocity with which the ball was thrown up (ii) the maximum height attained by the ball and (iii) the distance of the ball below the highest point after \(2\sec \). Take, \(g = 10m.{\sec ^{ - 2}}\)
Ans:
(51) A body is released from rest from a height. Find its speed at (i) \(t = 1\sec \) (ii) \(t = 2\sec \) (iii) \(t = 3\sec \) after the release.
Ans: 9.8 m/s, 19.6 m/s, 29.4 m/s
(52) A particle is released from rest from a height. Find the distance it falls through in (i) \(1\sec \) (ii) \(2\sec \) (iii) \(3\sec \)
Ans: 4.9 m, 19.6 m, 44.1 m
(53) A person takes out a coin from his pocket and drops it from a height of 1.6 m. With what speed does it strike the ground?
Ans: 5.6 m/s
(54) A body is thrown upwards with a speed of 19.6 m/s. Find its speed after (i) 1 sec (ii) 2 sec
Ans: 9.8 m/s, 0 m/s
(55) A body is thrown upwards with a speed of 29.4 m/s. Find the height the particle rises through in (i) 1 sec and (ii) 2 sec.
Ans: 24.5 m, 39.2 m
(56) A particle is thrown upwards with a speed of 39.2 m/s. Find (i) the time for which it moves in the upward direction, and (ii) the maximum height reached.
Ans: 4 sec, 78.4 m
(57) A ball is thrown upwards with some initial speed. It goes up to a height of 19.6 m and then returns. Find (i) the initial speed, (ii) the time taken in reaching the highest point, (iii) the velocity of the ball one second before and one second after it reaches the maximum height, and (iv) the time taken by the ball to return to its original position.
Ans: \(19.6m/\sec \), \(2\sec \), \(9.8m/s\), \( - 9.8m/\sec \), \(4\sec \)
(58) A body is dropped from some height. It moves through a distance of \(24.5m\) in the last second before hitting the ground. Find the height from which it was dropped.
Ans: 44.1 metres
(59) A ball is dropped from the edge of a roof. It takes 0.1 sec to crossed a window of height 2.0 m. Find the height of the roof above the top of the windows.
Ans: 19.6 metres
(60) A ball is dropped from a height of 20 m. At the same instant another ball is thrown up from the ground with a speed of 20 m/sec. When and where will the balls meet?
Ans: 1 sec, 15.1 m from above the ground.
(61) A ball 'A' is dropped from a 44.1 metre high cliff. Two seconds later, another ball 'B' is thrown downwards from the same place with some initial speed. The two balls reach the ground together. Find the speed with which the ball 'B' wae thrown.
Ans: 39.2 m/sec
(62) A ball is dropped from a cliff. Find (i) its speed 2 sec afterit is dropped, (ii) its speed when it has fallen through 78.4 m and (iii) the time taken in falling through 78.4 m
Ans: 19.6 m, 39.2 m, 4 sec
(63) A ball is thrown upwards with a speed of 39.2 m/sec. Calculate (i) the maximum height it reaches and (ii) the time taken in reaching the maximum height.
And: 78.4 m, 4 sec
(64) A ball is thrown upwards takes 4 sec to reach the maximum height. Find (i) the initial speed with which it was thrown and (ii) the maximum height reached.
Ans: 39.2 m, 78.4 m
(65) An object thrown upwards reaches the highest point in 5 sec. Find the velocity with which it was thrown.
Ans: 49 m/s
(66) A stone thrown upwards attain a maximum height of 19.6 m. Find the velocity with which it was thrown.
Ans: 19.6 m/sec
(67) A body is thrown upwards with a velocity of 20 m/s. How much time will it take to return to its original position?
Ans: 4.08 sec
(68) A ball is dropped from a height 2.50 m above the floor. (i) Find the speed 'v' with which it reaches the floor. (ii) The ball now rebounds. The speed of the ball is decreased to \(\frac{{3v}}{4}\) due to this collision. How high will the ball rise?
Ans: 7 m/s, 1.4 m
(69) A stone is dropped from a cliff at 22:30:30 pm . Another stone is dropped from the same point at 2:30:31 p.m. Find the separation between the stone at (i) 2:30:31 p.m (ii) 2:30:35 p.m.
Ans: 4.9 m, 44.1 m
(70) A ball is thrown upwards from the surface of the moon with a velocity of 19.6 m/s. (i) How much time will it take to attain the maximum height? (ii) How high will it go?
Ans: 12 sec, 117.6 m
(71) A flowerpot drops from the edge of the roof of a multi storied building. Calculate the time taken by the pot to cross a particular distance AB of height 2.9 m, the upper point A being 19.6 m below the roof.
Ans: \(\frac{1}{7}\sec \)
(72) A wicketkeeping glove is dropped from a height of 40 m and simultaneously a ball is thrown upwards from the ground with a speed of 40 m/sec. When and where do thwy meet?
Ans: 1 sec after the glove is dropped and 35.1 m above the ground.
(73) A boy on a 78.4 m high cliff drops a stone. One second later, he throws another stone downwards with some speed. The two stone reach the ground simultaneously. Find two stone with which the ground simultaneously. Find the speed with which the second stone was thrown.
Ans: 11.43 m/s
Mass & Weight
(74) Find the weight of a baby of mass 5 kg.
Ans: 49 N
(75) What is the mass of an object whose weight is \(49N\)?
Ans: \(5kg\)
(76) The weight of a body is \(50N\). What is its mass? Taking \(g = 9.8m.{\sec ^{ - 2}}\).
Ans:
(77) How much would a \(70kg\) man weight on the moon? What would be his mass on the earth and on the moon? Acceleration due to gravity on moon is \(1.63m.{\sec ^{ - 2}}\)
Ans:
(1) এই সাইটের সবকিছু Content বিনামুল্যে, এই সাইট থেকে নোট্স, সাজেশন এসব কোনও কিছুর জন্য কোনোরূপ মূল্য ধরা হয় না।
(2) বাংলাতে লেখা এরকম কোনো ওয়েবসাইট/ব্লগ নেই যেখান থেকে ছাত্র/ছাত্রীরা তাদের পড়াশোনা সংক্রান্ত বিষয়ে উপকৃত হতে পারে, তাই এটা আমার ঐকান্তিক প্রচেষ্টায় তৈরি করেছি।
(3) নবম, দশম, একাদশ, দ্বাদশ শ্রেণির ছাত্ররা এখানে তাদের বিষয়বস্তু গুলিকে Animation, Video, Java Applet এর মাধ্যমে খুব সহজে বুঝতে পারবে।
(4) এই ওয়েবসাইটে ভিজিট করার পর যদি মনে করো সত্যি এটি একটা Helpful Blog Site তাহলে যেকোনো Post গুলির Comment Section এ গিয়ে Comment দাও। এতে আমি উৎসাহ পাই, আর তোমাদের দেওয়া উৎসাহ আমি এটাকে একদিন হয়তো Complete Educational Guide Website হিসাবে তোমাদের কাছে তুলে ধরতে পারবো।
(5) এই Website থেকে তুমি যতটা উপকৃত হবে, সম্পূর্রূপে বিনামূল্যে। কিন্তু এই http://hemanta.org নামটি কিনতে বছরে প্রায় 2500 টাকা দিতে হয়। আমার লেখার Content গুলো টাইপ করার জন্য কিছু খরচ করতে হয়। আর এই সমস্ত খরচটা উঠে আসে এই Website এ বসানো Advertisement গুলি থেকে।
(6) তাই সবাইকে অনুরোধ যদি মনে করো এই সাইটে এসে আমি উপকৃত হয়েছি তার বিনিময়ে তোমরা এই Website এ বসানো Advertisement গুলিতে যতবার পারো Click দাও। তোমার একটা Click এর মূল্য 0.001$ অর্থাৎ প্রায় 0.06 পয়সা। এতে আমার গাড়ী, বাড়ী হবে না, কিন্তু বছরের শেষে হয়তো এই Website এর কিছুটা খরচ হয়তো লাঘব হবে। এটা তোমাদেরকে মানবিক ভাবে অনুরোধ রাখলাম।
(7) নতুন নতুন পোষ্ট গুলো লেখা চলছে, তাই আমাদের ফেসবুক পেজ Millennium Physics এ লাইক দিয়ে রাখো এবং আমাদের Google Plus Account এ যুক্ত হও। নতুন পোষ্ট দেওয়া মাত্রই তুমি পেয়ে যাবে।
(8) আরও অন্য কিছু বিষয় জানার প্রয়োজন থাকলে আমাদের Website এর একদম শেষে Contact Form আছে। যোগাযোগ করতে পারো।
(9) আপাতত ক্লাস IX, X, XI, XII, B. Sc Nursing, WBJEE, NEET এর Physics নিয়ে লেখা চলছে। আমাদের সঙ্গে থাকার জন্য অনুরোধ রইল।
COMMENTS