16. A cell can be balanced against 110 cm and 100 cm of potentiometer wire, respectively with and without being short circuited through a resistance of 10Ω. Its internal resistance is
    

1. 1. ​1.0 ohm ​

   
   
   

   2. 0.5 ohm ​

   
   
   

   3. 2.0 ohm

   
   
   

   4. zero

   
   
   

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1. View Hint View Answer Discuss in Forum

   Here

   E >	ER
   	R + r

   
   hence the lengths 110 cm and 100 cm are interchanged.
   Without being short-circuited through R, only the battery E is balanced.
   

   E =	V	× l1 =	V	× 110 ................(i)
   	L		L

   
   

   When R is connected across E, Ri =	V	× l2
   	L

   
   

   Or, R		E		=	V	× 100..............(ii)
   		R + r			L

   
   Dividing (i) by (ii), we get
   

   	R + r	=	110
   	R		100

   
   or, 100 R + 100 r = 110 R

   Correct Option: A

   Here

   E >	ER
   	R + r

   
   hence the lengths 110 cm and 100 cm are interchanged.
   Without being short-circuited through R, only the battery E is balanced.
   

   E =	V	× l1 =	V	× 110 ................(i)
   	L		L

   
   

   When R is connected across E, Ri =	V	× l2
   	L

   
   

   Or, R		E		=	V	× 100..............(ii)
   		R + r			L

   
   Dividing (i) by (ii), we get
   

   	R + r	=	110
   	R		100

   
   or, 100 R + 100 r = 110 R

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17. A steady current of 1.5 amp flows through a copper voltameter for 10 minutes. If the electrochemical equivalent of copper is 30 × 10^–5 g coulomb^–1, the mass of copper deposited on the electrode will be​

1. 1. 0.50 g

   
   
   

   2. 0.67 g ​

   
   
   

   3. 0.27 g ​

   
   
   

   4. 0.40 g.

   
   
   

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1. View Hint View Answer Discuss in Forum

   We have,  m = ZIt
   ​where, Z is the electrochemical equivalent of copper. ​
   ⇒ m = 30 x 10^-5 x 1.5 x 10 x 60
   ​= 0.27 gm.

   Correct Option: C

   We have,  m = ZIt
   ​where, Z is the electrochemical equivalent of copper. ​
   ⇒ m = 30 x 10^-5 x 1.5 x 10 x 60
   ​= 0.27 gm.

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18. See the electric circuit shown in the figure.
    
    Which of the following equations is a correct equation for it?​​

1. 1. ε₂ – i₂ r₂ – ε₁ – i₁ r₁ = 0 ​

   
   
   

   2. − ε₂ – (i₁ + i₂) R+ i₂ r₂ = 0 ​

   
   
   

   3. ε₁ – (i₁ + i₂) R + i₁ r₁ = 0 ​

   
   
   

   4. ε₁ – (i₁ + i₂) R– i₁ r₁ = 0
      

   
   
   

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1. View Hint View Answer Discuss in Forum

   
   Applying Kirchhoff ’s rule in loop abcfa
   ​ε₁ – (i₁ + i₂) R – i₁  r₁ = 0.
   

   Correct Option: D

   
   Applying Kirchhoff ’s rule in loop abcfa
   ​ε₁ – (i₁ + i₂) R – i₁  r₁ = 0.
   

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19. Kirchhoff’s first and second laws for electrical circuits are consequences of​

1. 1. conservation of electric charge and energy respectively ​

   
   
   

   2. conservation of electric charge ​

   
   
   

   3. conservation of energy and electric charge respectively ​

   
   
   

   4. conservation of energy

   
   
   

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1. View Hint View Answer Discuss in Forum

   ​Kirchhoff's first law deals with conservation of electrical charge & the second law deals with conservation of electrical energy.
   

   Correct Option: A

   ​Kirchhoff's first law deals with conservation of electrical charge & the second law deals with conservation of electrical energy.
   

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20. Two cells, having the same e.m.f., are connected in series through an external resistance R. Cells have internal resistances r₁ and r₂ (r₁ > r₂) respectively. When the circuit is closed, the potential difference across the first cell is zero. The value of R is​​​

1. 1. 	r1 + r2
      	2

   
   
   

   2. 	r1 - r2
      	2

   
   
   

   3. r₁ + r₂

   
   
   

   4. r₁ - r₂

   
   
   

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1. View Hint View Answer Discuss in Forum

   Current in the circuit
   

   =	E + E	=	2E
   	r1 + r2 + R		r1 + r2 + R

   
   P.D. across first cell = E – ir₁
   

   = E -	2E × r1	= 0
   	(r1 + r2) + R

   
   

   Now, E =	2Er1	= 0
   	(r1 + r2) + R

   
   

   ⇒ E =	2Er1	⇒ 2r1 = r1 + r2 + R
   	(r1 + r2) + R

   
   ⇒ R = r₁ - r₂

   Correct Option: D

   Current in the circuit
   

   =	E + E	=	2E
   	r1 + r2 + R		r1 + r2 + R

   
   P.D. across first cell = E – ir₁
   

   = E -	2E × r1	= 0
   	(r1 + r2) + R

   
   

   Now, E =	2Er1	= 0
   	(r1 + r2) + R

   
   

   ⇒ E =	2Er1	⇒ 2r1 = r1 + r2 + R
   	(r1 + r2) + R

   
   ⇒ R = r₁ - r₂

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