21. The density of material in CGS system of units is 4g/cm³. In a system of units in which unit of length is 10 cm and unit of mass is 100 g, the value of density of material will be​​
    

1. 1. 0.4

   
   
   

   2. 40

   
   
   

   3. 400

   
   
   

   4. 0.04

   
   
   

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1. View Hint View Answer Discuss in Forum

   In CGS system, ​ ​ d = 4	g
   	cm3

   
   The unit of mass is 100g and unit of length is 10 cm, so density
   

   =	4		100 g			=			4
   			100						100			(100 g)
   		10	cm		3				1		3	(10 cm)3
   		10							10

   
   

   =	4	× (10)3 .	100 g	= 40 unit
   	100		(10 cm)3

   Correct Option: B

   In CGS system, ​ ​ d = 4	g
   	cm3

   
   The unit of mass is 100g and unit of length is 10 cm, so density
   

   =	4		100 g			=			4
   			100						100			(100 g)
   		10	cm		3				1		3	(10 cm)3
   		10							10

   
   

   =	4	× (10)3 .	100 g	= 40 unit
   	100		(10 cm)3

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22. The frequency of vibration  f of a mass m suspended from a spring of spring constant k is given by a relation of the type f = c m^x k^y , where c is a dimensionless constant. The values of x and y are

1. 1. x =	1	, y =	1
      	2		2

   
   
   

   2. x = -	1	, y = -	1
      	2		2

   
   
   

   3. x =	1	, y = -	1
      	2		2

   
   
   

   4. x = -	1	, y =	1
      	2		2

   
   
   

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1. View Hint View Answer Discuss in Forum

   f = c m^x k^y ; ​
   Spring constant k = force / length.
   ​[M⁰ L⁰ T^{– 1}] = [M^x (MT^{– 2})^y] = [ M^{x + y} T^{– 2y}]
   

   ⇒ x - y = 0 , -2y = -1 or y =	1
   	2

   
   

   Therefore , x = -	1
   	2

   
   

   Correct Option: D

   f = c m^x k^y ; ​
   Spring constant k = force / length.
   ​[M⁰ L⁰ T^{– 1}] = [M^x (MT^{– 2})^y] = [ M^{x + y} T^{– 2y}]
   

   ⇒ x - y = 0 , -2y = -1 or y =	1
   	2

   
   

   Therefore , x = -	1
   	2

   
   

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23. In a vernier calliper N divisions of vernier scale coincides with (N – 1) divisions of main scale (in which length of one division is 1 mm). The least count of the instrument should be​​
    

1. 1. N

   
   
   

   2. N – 1​

   
   
   

   3. 1/10 N

   
   
   

   4. 1/N – 1

   
   
   

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1. View Hint View Answer Discuss in Forum

   ​Least count = 1MSD – 1 VSD
   

   = 1 MSD -		N - 1		MSD
   		N

   
   

   ( ∵ N VSD = (N - 1)MSD ∴ 1 VSD =	N - 1	MSD )
   	N

   
   

   =	1	MSD =	1	×	1	cm =	1
   	N		N		10		10 N

   Correct Option: C

   ​Least count = 1MSD – 1 VSD
   

   = 1 MSD -		N - 1		MSD
   		N

   
   

   ( ∵ N VSD = (N - 1)MSD ∴ 1 VSD =	N - 1	MSD )
   	N

   
   

   =	1	MSD =	1	×	1	cm =	1
   	N		N		10		10 N

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24. The density of a cube is measured by measuring its mass and length of its sides. If the maximum error in the measurement of mass and length are 4% and 3% respectively, the maximum error in the measurement of density will be
    

1. 1. 7%

   
   
   

   2. ​9%

   
   
   

   3. 12%

   
   
   

   4. 13%

   
   
   

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1. View Hint View Answer Discuss in Forum

   Density =	Mass
   	Volume

   
   

   ρ =	M
   	L3

   
   

   ∴	∆ρ	=	∆M	- 3	∆L
   	ρ		M		L

   
   % error in density = % error in Mass ​​​​ + 3 (% error in length) ​ 
   = 4 + 3(3) = 13%
   

   Correct Option: D

   Density =	Mass
   	Volume

   
   

   ρ =	M
   	L3

   
   

   ∴	∆ρ	=	∆M	- 3	∆L
   	ρ		M		L

   
   % error in density = % error in Mass ​​​​ + 3 (% error in length) ​ 
   = 4 + 3(3) = 13%
   

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25. If the error in the measurement of radius of a sphere is 2%, then the error in the determination of volume of the sphere will be :​
    

1. 1. 4%

   
   
   

   2. ​6%

   
   
   

   3. 8%

   
   
   

   4. ​2%

   
   
   

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1. View Hint View Answer Discuss in Forum

   Error in the measurement of radius of a sphere = 2%
   

   Volume of the sphere =	4	πr3
   	3

   
   

   ∴ Error in the volume = 3.	∆r	= 3 × 2% = 6%
   	r

   Correct Option: B

   Error in the measurement of radius of a sphere = 2%
   

   Volume of the sphere =	4	πr3
   	3

   
   

   ∴ Error in the volume = 3.	∆r	= 3 × 2% = 6%
   	r

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