QBManager

Current Electricity

152854 In balanced metre bridge $5 Ω$ in the left gap and $R Ω$ in the right gap. When $R Ω$ is shunted with an equal resistance, the new balance point is at $1.6 l_{1}$ where ' $l_{1}$ ' is the earlier balancing length. The value of ' $l 1$ ' is

1

25 cm

2

35 cm

3

30 cm

4

40 cm

Explanation:

A
At balance point -
$\frac{5}{R} = \frac{l_{1}}{100 - l_{1}}$
In the second case-

$\frac{5}{R / 2} = \frac{1.6 l_{1}}{100 - 1.6 l_{1}}$
$\frac{10}{R} = \frac{1.6 l_{1}}{100 - 1.6 l_{1}}$
Dividing equation (i) by (ii),
$\frac{5}{10} = \frac{l_{1}}{100 - l_{1}} \times \frac{100 - 1.6 l_{1}}{1.6 l_{1}}$
$\frac{1}{2} = \frac{100 - 1.6 l_{1}}{(100 - l_{1}) \times 1.6}$
$200 - 3.2 l_{1} = 160 - 1.6 l_{1}$
$1.6 l_{1} = 40$
$∴ l_{1} = \frac{40}{1.6} = 25 cm$

MHT-CET 2020

Current Electricity

152855 The given circuit is balanced Wheatstone's bridge. The value of resistance ' $x$ ' is

1

12 Ω

2

4 Ω

3

6 Ω

4

24 Ω

Explanation:

A
Point B \& D is at same potential

At Wheatstone's bridge
$\frac{8}{4 + x} = \frac{1}{2}$
$16 = 4 + x$
$x = 12 Ω$

MHT-CET 2020

Current Electricity

152856 A galvanometer of resistance $20 Ω$ has a current sensitivity of $5 div / mA$ . The instrument has 50 divisions. It can be converted into a voltmeter reading upto 25 volt by connecting a resistance of

1

2480 Ω

in series

2

20 Ω

in parallel

3

1240 Ω

in series

4

2480 Ω

in parallel

Explanation:

A Given that,
$G = 20 Ω$
$V = 25 V$
Current sensitivity $= 5 div / mA$
$∴$ For 50 division, the current
$I_{g} = 10 mA = 10 \times 10^{- 3} A$
To convert the galvanometer into a voltmeter, a high resistance must be connected in series with the galvanometer coil. This series multiplies
$R_{S} = \frac{V}{I_{g}} - G = \frac{25}{10 \times 10^{- 3}} - 20 = 2500 - 20$
$= 2480 Ω in series$

MHT-CET 2020

Current Electricity

152857 In a Wheatstone's bridge, the resistance in the three arms are $P, Q, R$ and its fourth arm has a parallel combination of two resistances $S_{1}$ and $S_{2}$ , The balancing condition of the bridge is

1

\frac{P}{Q} = \frac{2 R}{S_{1} + S_{2}}

2

\frac{P}{Q} = \frac{R (S_{1} + S_{2})}{2 S_{1} S_{2}}

3

\frac{P}{Q} = (\frac{R S_{1} S_{2}}{S_{1} + S_{2}})

4

\frac{P}{Q} = \frac{R (S_{1} + S_{2})}{S_{1} S_{2}}

Explanation:

D The given figure is equivalent to-

For balanced bridge,
$\frac{P}{R} = \frac{Q}{\frac{S_{1} S_{2}}{S_{1} + S_{2}}}$
$\frac{P}{R} = \frac{Q (S_{1} + S_{2})}{S_{1} S_{2}}$
$\frac{P}{Q} = \frac{R (S_{1} + S_{2})}{S_{1} S_{2}}$

MHT-CET 2020

Current Electricity

152859 In potentiometer experiment, null point is obtained at a particular point on the potentiometer wire for a cell. If the length of the potentiometer wire is increased without changing the driving cell, the balancing length will.

1 remains same

2 increase

3 become zero

4 decrease

Explanation:

B Case I : $- E = \frac{V}{L} x$

Case II :- Driving cell V remains constant
$E = \frac{V}{L^{'}} y$

From equation (i), (ii), we get -
$\frac{V}{L} x = \frac{V}{L^{'}} y$
$y = \frac{L^{'}}{L} x$
$∵ L^{'} > L$
$∴ y > x$
Hence, balancing length will increase.

MHT-CET 2020

Current Electricity

152854 In balanced metre bridge $5 Ω$ in the left gap and $R Ω$ in the right gap. When $R Ω$ is shunted with an equal resistance, the new balance point is at $1.6 l_{1}$ where ' $l_{1}$ ' is the earlier balancing length. The value of ' $l 1$ ' is

1

25 cm

2

35 cm

3

30 cm

4

40 cm

Explanation:

A
At balance point -
$\frac{5}{R} = \frac{l_{1}}{100 - l_{1}}$
In the second case-

$\frac{5}{R / 2} = \frac{1.6 l_{1}}{100 - 1.6 l_{1}}$
$\frac{10}{R} = \frac{1.6 l_{1}}{100 - 1.6 l_{1}}$
Dividing equation (i) by (ii),
$\frac{5}{10} = \frac{l_{1}}{100 - l_{1}} \times \frac{100 - 1.6 l_{1}}{1.6 l_{1}}$
$\frac{1}{2} = \frac{100 - 1.6 l_{1}}{(100 - l_{1}) \times 1.6}$
$200 - 3.2 l_{1} = 160 - 1.6 l_{1}$
$1.6 l_{1} = 40$
$∴ l_{1} = \frac{40}{1.6} = 25 cm$

MHT-CET 2020

Current Electricity

152855 The given circuit is balanced Wheatstone's bridge. The value of resistance ' $x$ ' is

1

12 Ω

2

4 Ω

3

6 Ω

4

24 Ω

Explanation:

A
Point B \& D is at same potential

At Wheatstone's bridge
$\frac{8}{4 + x} = \frac{1}{2}$
$16 = 4 + x$
$x = 12 Ω$

MHT-CET 2020

Current Electricity

152856 A galvanometer of resistance $20 Ω$ has a current sensitivity of $5 div / mA$ . The instrument has 50 divisions. It can be converted into a voltmeter reading upto 25 volt by connecting a resistance of

1

2480 Ω

in series

2

20 Ω

in parallel

3

1240 Ω

in series

4

2480 Ω

in parallel

Explanation:

A Given that,
$G = 20 Ω$
$V = 25 V$
Current sensitivity $= 5 div / mA$
$∴$ For 50 division, the current
$I_{g} = 10 mA = 10 \times 10^{- 3} A$
To convert the galvanometer into a voltmeter, a high resistance must be connected in series with the galvanometer coil. This series multiplies
$R_{S} = \frac{V}{I_{g}} - G = \frac{25}{10 \times 10^{- 3}} - 20 = 2500 - 20$
$= 2480 Ω in series$

MHT-CET 2020

Current Electricity

152857 In a Wheatstone's bridge, the resistance in the three arms are $P, Q, R$ and its fourth arm has a parallel combination of two resistances $S_{1}$ and $S_{2}$ , The balancing condition of the bridge is

1

\frac{P}{Q} = \frac{2 R}{S_{1} + S_{2}}

2

\frac{P}{Q} = \frac{R (S_{1} + S_{2})}{2 S_{1} S_{2}}

3

\frac{P}{Q} = (\frac{R S_{1} S_{2}}{S_{1} + S_{2}})

4

\frac{P}{Q} = \frac{R (S_{1} + S_{2})}{S_{1} S_{2}}

Explanation:

D The given figure is equivalent to-

For balanced bridge,
$\frac{P}{R} = \frac{Q}{\frac{S_{1} S_{2}}{S_{1} + S_{2}}}$
$\frac{P}{R} = \frac{Q (S_{1} + S_{2})}{S_{1} S_{2}}$
$\frac{P}{Q} = \frac{R (S_{1} + S_{2})}{S_{1} S_{2}}$

MHT-CET 2020

Current Electricity

152859 In potentiometer experiment, null point is obtained at a particular point on the potentiometer wire for a cell. If the length of the potentiometer wire is increased without changing the driving cell, the balancing length will.

1 remains same

2 increase

3 become zero

4 decrease

Explanation:

B Case I : $- E = \frac{V}{L} x$

Case II :- Driving cell V remains constant
$E = \frac{V}{L^{'}} y$

From equation (i), (ii), we get -
$\frac{V}{L} x = \frac{V}{L^{'}} y$
$y = \frac{L^{'}}{L} x$
$∵ L^{'} > L$
$∴ y > x$
Hence, balancing length will increase.

MHT-CET 2020

Current Electricity

152854 In balanced metre bridge $5 Ω$ in the left gap and $R Ω$ in the right gap. When $R Ω$ is shunted with an equal resistance, the new balance point is at $1.6 l_{1}$ where ' $l_{1}$ ' is the earlier balancing length. The value of ' $l 1$ ' is

1

25 cm

2

35 cm

3

30 cm

4

40 cm

Explanation:

A
At balance point -
$\frac{5}{R} = \frac{l_{1}}{100 - l_{1}}$
In the second case-

$\frac{5}{R / 2} = \frac{1.6 l_{1}}{100 - 1.6 l_{1}}$
$\frac{10}{R} = \frac{1.6 l_{1}}{100 - 1.6 l_{1}}$
Dividing equation (i) by (ii),
$\frac{5}{10} = \frac{l_{1}}{100 - l_{1}} \times \frac{100 - 1.6 l_{1}}{1.6 l_{1}}$
$\frac{1}{2} = \frac{100 - 1.6 l_{1}}{(100 - l_{1}) \times 1.6}$
$200 - 3.2 l_{1} = 160 - 1.6 l_{1}$
$1.6 l_{1} = 40$
$∴ l_{1} = \frac{40}{1.6} = 25 cm$

MHT-CET 2020

Current Electricity

152855 The given circuit is balanced Wheatstone's bridge. The value of resistance ' $x$ ' is

1

12 Ω

2

4 Ω

3

6 Ω

4

24 Ω

Explanation:

A
Point B \& D is at same potential

At Wheatstone's bridge
$\frac{8}{4 + x} = \frac{1}{2}$
$16 = 4 + x$
$x = 12 Ω$

MHT-CET 2020

Current Electricity

152856 A galvanometer of resistance $20 Ω$ has a current sensitivity of $5 div / mA$ . The instrument has 50 divisions. It can be converted into a voltmeter reading upto 25 volt by connecting a resistance of

1

2480 Ω

in series

2

20 Ω

in parallel

3

1240 Ω

in series

4

2480 Ω

in parallel

Explanation:

A Given that,
$G = 20 Ω$
$V = 25 V$
Current sensitivity $= 5 div / mA$
$∴$ For 50 division, the current
$I_{g} = 10 mA = 10 \times 10^{- 3} A$
To convert the galvanometer into a voltmeter, a high resistance must be connected in series with the galvanometer coil. This series multiplies
$R_{S} = \frac{V}{I_{g}} - G = \frac{25}{10 \times 10^{- 3}} - 20 = 2500 - 20$
$= 2480 Ω in series$

MHT-CET 2020

Current Electricity

152857 In a Wheatstone's bridge, the resistance in the three arms are $P, Q, R$ and its fourth arm has a parallel combination of two resistances $S_{1}$ and $S_{2}$ , The balancing condition of the bridge is

1

\frac{P}{Q} = \frac{2 R}{S_{1} + S_{2}}

2

\frac{P}{Q} = \frac{R (S_{1} + S_{2})}{2 S_{1} S_{2}}

3

\frac{P}{Q} = (\frac{R S_{1} S_{2}}{S_{1} + S_{2}})

4

\frac{P}{Q} = \frac{R (S_{1} + S_{2})}{S_{1} S_{2}}

Explanation:

D The given figure is equivalent to-

For balanced bridge,
$\frac{P}{R} = \frac{Q}{\frac{S_{1} S_{2}}{S_{1} + S_{2}}}$
$\frac{P}{R} = \frac{Q (S_{1} + S_{2})}{S_{1} S_{2}}$
$\frac{P}{Q} = \frac{R (S_{1} + S_{2})}{S_{1} S_{2}}$

MHT-CET 2020

Current Electricity

152859 In potentiometer experiment, null point is obtained at a particular point on the potentiometer wire for a cell. If the length of the potentiometer wire is increased without changing the driving cell, the balancing length will.

1 remains same

2 increase

3 become zero

4 decrease

Explanation:

B Case I : $- E = \frac{V}{L} x$

Case II :- Driving cell V remains constant
$E = \frac{V}{L^{'}} y$

From equation (i), (ii), we get -
$\frac{V}{L} x = \frac{V}{L^{'}} y$
$y = \frac{L^{'}}{L} x$
$∵ L^{'} > L$
$∴ y > x$
Hence, balancing length will increase.

MHT-CET 2020

Current Electricity

152854 In balanced metre bridge $5 Ω$ in the left gap and $R Ω$ in the right gap. When $R Ω$ is shunted with an equal resistance, the new balance point is at $1.6 l_{1}$ where ' $l_{1}$ ' is the earlier balancing length. The value of ' $l 1$ ' is

1

25 cm

2

35 cm

3

30 cm

4

40 cm

Explanation:

A
At balance point -
$\frac{5}{R} = \frac{l_{1}}{100 - l_{1}}$
In the second case-

$\frac{5}{R / 2} = \frac{1.6 l_{1}}{100 - 1.6 l_{1}}$
$\frac{10}{R} = \frac{1.6 l_{1}}{100 - 1.6 l_{1}}$
Dividing equation (i) by (ii),
$\frac{5}{10} = \frac{l_{1}}{100 - l_{1}} \times \frac{100 - 1.6 l_{1}}{1.6 l_{1}}$
$\frac{1}{2} = \frac{100 - 1.6 l_{1}}{(100 - l_{1}) \times 1.6}$
$200 - 3.2 l_{1} = 160 - 1.6 l_{1}$
$1.6 l_{1} = 40$
$∴ l_{1} = \frac{40}{1.6} = 25 cm$

MHT-CET 2020

Current Electricity

152855 The given circuit is balanced Wheatstone's bridge. The value of resistance ' $x$ ' is

1

12 Ω

2

4 Ω

3

6 Ω

4

24 Ω

Explanation:

A
Point B \& D is at same potential

At Wheatstone's bridge
$\frac{8}{4 + x} = \frac{1}{2}$
$16 = 4 + x$
$x = 12 Ω$

MHT-CET 2020

Current Electricity

152856 A galvanometer of resistance $20 Ω$ has a current sensitivity of $5 div / mA$ . The instrument has 50 divisions. It can be converted into a voltmeter reading upto 25 volt by connecting a resistance of

1

2480 Ω

in series

2

20 Ω

in parallel

3

1240 Ω

in series

4

2480 Ω

in parallel

Explanation:

A Given that,
$G = 20 Ω$
$V = 25 V$
Current sensitivity $= 5 div / mA$
$∴$ For 50 division, the current
$I_{g} = 10 mA = 10 \times 10^{- 3} A$
To convert the galvanometer into a voltmeter, a high resistance must be connected in series with the galvanometer coil. This series multiplies
$R_{S} = \frac{V}{I_{g}} - G = \frac{25}{10 \times 10^{- 3}} - 20 = 2500 - 20$
$= 2480 Ω in series$

MHT-CET 2020

Current Electricity

152857 In a Wheatstone's bridge, the resistance in the three arms are $P, Q, R$ and its fourth arm has a parallel combination of two resistances $S_{1}$ and $S_{2}$ , The balancing condition of the bridge is

1

\frac{P}{Q} = \frac{2 R}{S_{1} + S_{2}}

2

\frac{P}{Q} = \frac{R (S_{1} + S_{2})}{2 S_{1} S_{2}}

3

\frac{P}{Q} = (\frac{R S_{1} S_{2}}{S_{1} + S_{2}})

4

\frac{P}{Q} = \frac{R (S_{1} + S_{2})}{S_{1} S_{2}}

Explanation:

D The given figure is equivalent to-

For balanced bridge,
$\frac{P}{R} = \frac{Q}{\frac{S_{1} S_{2}}{S_{1} + S_{2}}}$
$\frac{P}{R} = \frac{Q (S_{1} + S_{2})}{S_{1} S_{2}}$
$\frac{P}{Q} = \frac{R (S_{1} + S_{2})}{S_{1} S_{2}}$

MHT-CET 2020

Current Electricity

152859 In potentiometer experiment, null point is obtained at a particular point on the potentiometer wire for a cell. If the length of the potentiometer wire is increased without changing the driving cell, the balancing length will.

1 remains same

2 increase

3 become zero

4 decrease

Explanation:

B Case I : $- E = \frac{V}{L} x$

Case II :- Driving cell V remains constant
$E = \frac{V}{L^{'}} y$

From equation (i), (ii), we get -
$\frac{V}{L} x = \frac{V}{L^{'}} y$
$y = \frac{L^{'}}{L} x$
$∵ L^{'} > L$
$∴ y > x$
Hence, balancing length will increase.

MHT-CET 2020

Current Electricity

152854 In balanced metre bridge $5 Ω$ in the left gap and $R Ω$ in the right gap. When $R Ω$ is shunted with an equal resistance, the new balance point is at $1.6 l_{1}$ where ' $l_{1}$ ' is the earlier balancing length. The value of ' $l 1$ ' is

1

25 cm

2

35 cm

3

30 cm

4

40 cm

Explanation:

A
At balance point -
$\frac{5}{R} = \frac{l_{1}}{100 - l_{1}}$
In the second case-

$\frac{5}{R / 2} = \frac{1.6 l_{1}}{100 - 1.6 l_{1}}$
$\frac{10}{R} = \frac{1.6 l_{1}}{100 - 1.6 l_{1}}$
Dividing equation (i) by (ii),
$\frac{5}{10} = \frac{l_{1}}{100 - l_{1}} \times \frac{100 - 1.6 l_{1}}{1.6 l_{1}}$
$\frac{1}{2} = \frac{100 - 1.6 l_{1}}{(100 - l_{1}) \times 1.6}$
$200 - 3.2 l_{1} = 160 - 1.6 l_{1}$
$1.6 l_{1} = 40$
$∴ l_{1} = \frac{40}{1.6} = 25 cm$

MHT-CET 2020

Current Electricity

152855 The given circuit is balanced Wheatstone's bridge. The value of resistance ' $x$ ' is

1

12 Ω

2

4 Ω

3

6 Ω

4

24 Ω

Explanation:

A
Point B \& D is at same potential

At Wheatstone's bridge
$\frac{8}{4 + x} = \frac{1}{2}$
$16 = 4 + x$
$x = 12 Ω$

MHT-CET 2020

Current Electricity

152856 A galvanometer of resistance $20 Ω$ has a current sensitivity of $5 div / mA$ . The instrument has 50 divisions. It can be converted into a voltmeter reading upto 25 volt by connecting a resistance of

1

2480 Ω

in series

2

20 Ω

in parallel

3

1240 Ω

in series

4

2480 Ω

in parallel

Explanation:

A Given that,
$G = 20 Ω$
$V = 25 V$
Current sensitivity $= 5 div / mA$
$∴$ For 50 division, the current
$I_{g} = 10 mA = 10 \times 10^{- 3} A$
To convert the galvanometer into a voltmeter, a high resistance must be connected in series with the galvanometer coil. This series multiplies
$R_{S} = \frac{V}{I_{g}} - G = \frac{25}{10 \times 10^{- 3}} - 20 = 2500 - 20$
$= 2480 Ω in series$

MHT-CET 2020

Current Electricity

152857 In a Wheatstone's bridge, the resistance in the three arms are $P, Q, R$ and its fourth arm has a parallel combination of two resistances $S_{1}$ and $S_{2}$ , The balancing condition of the bridge is

1

\frac{P}{Q} = \frac{2 R}{S_{1} + S_{2}}

2

\frac{P}{Q} = \frac{R (S_{1} + S_{2})}{2 S_{1} S_{2}}

3

\frac{P}{Q} = (\frac{R S_{1} S_{2}}{S_{1} + S_{2}})

4

\frac{P}{Q} = \frac{R (S_{1} + S_{2})}{S_{1} S_{2}}

Explanation:

D The given figure is equivalent to-

For balanced bridge,
$\frac{P}{R} = \frac{Q}{\frac{S_{1} S_{2}}{S_{1} + S_{2}}}$
$\frac{P}{R} = \frac{Q (S_{1} + S_{2})}{S_{1} S_{2}}$
$\frac{P}{Q} = \frac{R (S_{1} + S_{2})}{S_{1} S_{2}}$

MHT-CET 2020

Current Electricity

152859 In potentiometer experiment, null point is obtained at a particular point on the potentiometer wire for a cell. If the length of the potentiometer wire is increased without changing the driving cell, the balancing length will.

1 remains same

2 increase

3 become zero

4 decrease

Explanation:

B Case I : $- E = \frac{V}{L} x$

Case II :- Driving cell V remains constant
$E = \frac{V}{L^{'}} y$

From equation (i), (ii), we get -
$\frac{V}{L} x = \frac{V}{L^{'}} y$
$y = \frac{L^{'}}{L} x$
$∵ L^{'} > L$
$∴ y > x$
Hence, balancing length will increase.

MHT-CET 2020

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