Kitabcd Academy

Online Educational Study Material and PDF - Email : kitabcd2018@gmail.com

Numerical MCQ-1-Test–Class-10-Science-1-Chapter-1-Gravitation-MSBSHSE

#, #
  • If Teacher ask to give test then non-logged in student should enter email ID provided by teacher at the beginning in email option.
  • Give details when asked in between this test so that teacher will know who is given this test.
  • You can download solution of this MCQ after the test.
  • You will also get solution of this MCQ directly in your box  after submitting the test.
  • After submit wait for test result on the same page.

30 Mark Test 2-Mark-for each Question
Please provide email id (if you are not logged in) to get test result and answers directly in your inbox.

Please enter your email:

2-Mark – Solve numerical and select the correct option.

Take (G = 6.67 x 10-11 N-m2/kg2, g =9.8 m/s2)

Question-1 :

The time taken by the earth to complete one revolution around the Sun is 3.156 x 107 s. The distance between the earth and the Sun is 1.5 x 1011 m. Find the speed of revolution of the earth.

 
 
 
 

2-Mark – Solve numerical and select the correct option.

Take (G = 6.67 x 10−11 N-m2/kg2, g =9.8 m/s2)

Question-2 :

Assuming that the earth performs uniform circular motion around the Sun. What will be the centripetal acceleration of the earth ?

[Speed of the earth =3 x 104 m/s, distance between the earth and the Sun = 1.5 x 1011 m]

 
 
 
 

2-Mark – Solve numerical and select the correct option.

Take (G = 6.67 x 10−11 N-m2/kg2, g =9.8 m/s2)

Question-3 :

What will be the gravitational force on 60 kg man on the Moon?

M (Moon) = 7.36 x 1022 kg, R (Moon) = 1.74 x 106 m

 
 
 
 

2-Mark – Solve numerical and select the correct option.

Take (G = 6.67 x 10−11 N-m2/kg2, g =9.8 m/s2)

Question-4 :

What will be the gravitational force on 60 kg man on the Mars ?

M (Mars) = 6.4 x 1023 kg, R (Mars) = 3.395 x 106 m

 
 
 
 

2-Mark – Solve numerical and select the correct option.

Take (G = 6.67 x 10−11 N-m2/kg2, g =9.8 m/s2)

Question-5 :

Spheres A and B of uniform density have masses 1 kg and 100 kg respectively. Their centres are separated by 100 m. What will be the gravitational force between them?

 
 
 
 

2-Mark – Solve numerical and select the correct option.

Take (G = 6.67 x 10−11 N-m2/kg2, g =9.8 m/s2)

Question-6 :

Spheres A and B of uniform density have masses 1 kg and 100 kg respectively. Their centres are separated by 100 m. What will be the gravitational force on A due to earth.

[ M(earth) = 6 x 1024 kg, R(earth) = 6400 km]

 
 
 
 

2-Mark – Solve numerical and select the correct option.

Take (G = 6.67 x 10−11 N-m2/kg2, g =9.8 m/s2)

Question-7 :

Two spheres of uniform density have masses 10 kg and 40 kg. The distance between the centres of the spheres is 200 m. Find the gravitational force between them.

 
 
 
 

2-Mark – Solve numerical and select the correct option.

Take (G = 6.67 x 10−11 N-m2/kg2, g =9.8 m/s2)

Question-8 :

Find the gravitational force between a man of mass 50 kg and a car of mass 1500 kg separated by 10 m.

 
 
 
 

2-Mark – Solve numerical and select the correct option.

Take (G = 6.67 x 10−11 N-m2/kg2, g =9.8 m/s2)

Question-9 :

Find the magnitude of the gravitational force between the sun and the earth. (Mass of the Sun = 2 x 1030 kg, mass of the earth = 6 x 1024 kg and the distance between the centres of the Sun and the earth = 1.5 x 1011 m,)

 
 
 
 

2-Mark – Solve numerical and select the correct option.

Take (G = 6.67 x 10−11 N-m2/kg2, g =9.8 m/s2)

Question-10 :

The mass of the earth is 6 x 1024 kg. The distance between the earth and the Sun is 1.5 x 1011 m. If the gravitational force between the two is 3.5 x 1022 N, what is the mass of the Sun?

 
 
 
 

2-Mark – Solve numerical and select the correct option.

Take (G = 6.67 x 10−11 N-m2/kg2, g =9.8 m/s2)

Question-11 :

Find the magnitude of the acceleration due to gravity at the surface of the earth.

(M = 6 x 1024 kg, R = 6400 km)

 
 
 
 

2-Mark – Solve numerical and select the correct option.

Take (G = 6.67 x 10−11 N-m2/kg2, g =9.8 m/s2)

Question-12 :

If the acceleration due to gravity on the surface of the earth is 9.8 m/s2, what will be the acceleration due to gravity on the surface of a planet whose mass and radius both are two times the corresponding quantities for the earth?

 
 
 
 

2-Mark – Solve numerical and select the correct option.

Take (G = 6.67 x 10−11 N-m2/kg2, g =9.8 m/s2)

Question-13 :

An object thrown vertically upwards reaches a height, its initial velocity was 100 m/s. How long will the object take to come back to the earth? Assume g = 10 m/s2.

 
 
 
 

2-Mark – Solve numerical and select the correct option.

Take (G = 6.67 x 10−11 N-m2/kg2, g =9.8 m/s2)

Question-14 :

A body is released from the top of a building of height 19.6 m. Find the velocity with which the body hits the ground.

 
 
 
 

2-Mark – Solve numerical and select the correct option.

Take (G = 6.67 x 10−11 N-m2/kg2, g =9.8 m/s2)

Question-15 :

An object thrown vertically upwards reaches a height of 500 m. What was its initial velocity ? Assume g = 10 m/s2.

 
 
 
 

(*** Do not consider this as a question,  if shows wrong answer in result, ignore it.)

If your Teacher ask to attend this test then student should enter email ID provided by teacher at the beginning in email option and write self details as per below to know who is given the test.

Please enter

Your Full Name

Class Name

Roll No

Other details

Question 1 of 16

 

Related Post

Leave a Reply

Your email address will not be published. Required fields are marked *

Copyright © Kitabcd Academy-2024 ||-----------|| | BlogArise by Themeansar.

<<Menu
Kitabcd PDF Store
Tools