152854
In balanced metre bridge $5 \Omega$ in the left gap and $R \Omega$ in the right gap. When $R \Omega$ 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 $25 \mathrm{~cm}$
2 $35 \mathrm{~cm}$
3 $30 \mathrm{~cm}$
4 $40 \mathrm{~cm}$
Explanation:
A At balance point - $\frac{5}{\mathrm{R}}=\frac{l_{1}}{100-l_{1}}$ In the second case- $\frac{5}{\mathrm{R} / 2}=\frac{1.6 l_{1}}{100-1.6 l_{1}}$ $\frac{10}{\mathrm{R}}=\frac{1.6 l_{1}}{100-1.6 l_{1}}$ Dividing equation (i) by (ii), $\frac{5}{10}=\frac{l_{1}}{100-l_{1}} \times \frac{100-1.6 l_{1}}{1.6 l_{1}}$ $\frac{1}{2}=\frac{100-1.6 l_{1}}{\left(100-l_{1}\right) \times 1.6}$ $200-3.2 l_{1}=160-1.6 l_{1}$ $1.6 l_{1}=40$ $\therefore \quad l_{1}=\frac{40}{1.6}=25 \mathrm{~cm}$
MHT-CET 2020
Current Electricity
152855
The given circuit is balanced Wheatstone's bridge. The value of resistance ' $x$ ' is
1 $12 \Omega$
2 $4 \Omega$
3 $6 \Omega$
4 $24 \Omega$
Explanation:
A Point B \& D is at same potential At Wheatstone's bridge $\frac{8}{4+x}=\frac{1}{2}$ $16=4+x$ $x=12 \Omega$
MHT-CET 2020
Current Electricity
152856
A galvanometer of resistance $20 \Omega$ has a current sensitivity of $5 \mathrm{div} / \mathrm{mA}$. The instrument has 50 divisions. It can be converted into a voltmeter reading upto 25 volt by connecting a resistance of
1 $2480 \Omega$ in series
2 $20 \Omega$ in parallel
3 $1240 \Omega$ in series
4 $2480 \Omega$ in parallel
Explanation:
A Given that, $\mathrm{G}=20 \Omega$ $\mathrm{V}=25 \mathrm{~V}$ Current sensitivity $=5 \mathrm{div} / \mathrm{mA}$ $\therefore$ For 50 division, the current $\mathrm{I}_{\mathrm{g}}=10 \mathrm{~mA}=10 \times 10^{-3} \mathrm{~A}$ To convert the galvanometer into a voltmeter, a high resistance must be connected in series with the galvanometer coil. This series multiplies $\mathrm{R}_{\mathrm{S}}=\frac{\mathrm{V}}{\mathrm{I}_{\mathrm{g}}}-\mathrm{G}=\frac{25}{10 \times 10^{-3}}-20=2500-20$ $=2480 \Omega \text { in series }$
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
D The given figure is equivalent to- For balanced bridge, $\frac{\mathrm{P}}{\mathrm{R}}=\frac{\mathrm{Q}}{\frac{\mathrm{S}_{1} \mathrm{~S}_{2}}{\mathrm{~S}_{1}+\mathrm{S}_{2}}}$ $\frac{\mathrm{P}}{\mathrm{R}}=\frac{\mathrm{Q}\left(\mathrm{S}_{1}+\mathrm{S}_{2}\right)}{\mathrm{S}_{1} \mathrm{~S}_{2}}$ $\frac{\mathrm{P}}{\mathrm{Q}}=\frac{\mathrm{R}\left(\mathrm{S}_{1}+\mathrm{S}_{2}\right)}{\mathrm{S}_{1} \mathrm{~S}_{2}}$
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 : $-\mathrm{E}=\frac{\mathrm{V}}{\mathrm{L}} \mathrm{x}$ Case II :- Driving cell V remains constant $\mathrm{E}=\frac{\mathrm{V}}{\mathrm{L}^{\prime}} \mathrm{y}$ From equation (i), (ii), we get - $\frac{\mathrm{V}}{\mathrm{L}} \mathrm{x}=\frac{\mathrm{V}}{\mathrm{L}^{\prime}} \mathrm{y}$ $\mathrm{y}=\frac{\mathrm{L}^{\prime}}{\mathrm{L}} \mathrm{x}$ $\because \quad \mathrm{L}^{\prime}>\mathrm{L}$ $\therefore \quad \mathrm{y}>\mathrm{x}$ Hence, balancing length will increase.
152854
In balanced metre bridge $5 \Omega$ in the left gap and $R \Omega$ in the right gap. When $R \Omega$ 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 $25 \mathrm{~cm}$
2 $35 \mathrm{~cm}$
3 $30 \mathrm{~cm}$
4 $40 \mathrm{~cm}$
Explanation:
A At balance point - $\frac{5}{\mathrm{R}}=\frac{l_{1}}{100-l_{1}}$ In the second case- $\frac{5}{\mathrm{R} / 2}=\frac{1.6 l_{1}}{100-1.6 l_{1}}$ $\frac{10}{\mathrm{R}}=\frac{1.6 l_{1}}{100-1.6 l_{1}}$ Dividing equation (i) by (ii), $\frac{5}{10}=\frac{l_{1}}{100-l_{1}} \times \frac{100-1.6 l_{1}}{1.6 l_{1}}$ $\frac{1}{2}=\frac{100-1.6 l_{1}}{\left(100-l_{1}\right) \times 1.6}$ $200-3.2 l_{1}=160-1.6 l_{1}$ $1.6 l_{1}=40$ $\therefore \quad l_{1}=\frac{40}{1.6}=25 \mathrm{~cm}$
MHT-CET 2020
Current Electricity
152855
The given circuit is balanced Wheatstone's bridge. The value of resistance ' $x$ ' is
1 $12 \Omega$
2 $4 \Omega$
3 $6 \Omega$
4 $24 \Omega$
Explanation:
A Point B \& D is at same potential At Wheatstone's bridge $\frac{8}{4+x}=\frac{1}{2}$ $16=4+x$ $x=12 \Omega$
MHT-CET 2020
Current Electricity
152856
A galvanometer of resistance $20 \Omega$ has a current sensitivity of $5 \mathrm{div} / \mathrm{mA}$. The instrument has 50 divisions. It can be converted into a voltmeter reading upto 25 volt by connecting a resistance of
1 $2480 \Omega$ in series
2 $20 \Omega$ in parallel
3 $1240 \Omega$ in series
4 $2480 \Omega$ in parallel
Explanation:
A Given that, $\mathrm{G}=20 \Omega$ $\mathrm{V}=25 \mathrm{~V}$ Current sensitivity $=5 \mathrm{div} / \mathrm{mA}$ $\therefore$ For 50 division, the current $\mathrm{I}_{\mathrm{g}}=10 \mathrm{~mA}=10 \times 10^{-3} \mathrm{~A}$ To convert the galvanometer into a voltmeter, a high resistance must be connected in series with the galvanometer coil. This series multiplies $\mathrm{R}_{\mathrm{S}}=\frac{\mathrm{V}}{\mathrm{I}_{\mathrm{g}}}-\mathrm{G}=\frac{25}{10 \times 10^{-3}}-20=2500-20$ $=2480 \Omega \text { in series }$
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
D The given figure is equivalent to- For balanced bridge, $\frac{\mathrm{P}}{\mathrm{R}}=\frac{\mathrm{Q}}{\frac{\mathrm{S}_{1} \mathrm{~S}_{2}}{\mathrm{~S}_{1}+\mathrm{S}_{2}}}$ $\frac{\mathrm{P}}{\mathrm{R}}=\frac{\mathrm{Q}\left(\mathrm{S}_{1}+\mathrm{S}_{2}\right)}{\mathrm{S}_{1} \mathrm{~S}_{2}}$ $\frac{\mathrm{P}}{\mathrm{Q}}=\frac{\mathrm{R}\left(\mathrm{S}_{1}+\mathrm{S}_{2}\right)}{\mathrm{S}_{1} \mathrm{~S}_{2}}$
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 : $-\mathrm{E}=\frac{\mathrm{V}}{\mathrm{L}} \mathrm{x}$ Case II :- Driving cell V remains constant $\mathrm{E}=\frac{\mathrm{V}}{\mathrm{L}^{\prime}} \mathrm{y}$ From equation (i), (ii), we get - $\frac{\mathrm{V}}{\mathrm{L}} \mathrm{x}=\frac{\mathrm{V}}{\mathrm{L}^{\prime}} \mathrm{y}$ $\mathrm{y}=\frac{\mathrm{L}^{\prime}}{\mathrm{L}} \mathrm{x}$ $\because \quad \mathrm{L}^{\prime}>\mathrm{L}$ $\therefore \quad \mathrm{y}>\mathrm{x}$ Hence, balancing length will increase.
152854
In balanced metre bridge $5 \Omega$ in the left gap and $R \Omega$ in the right gap. When $R \Omega$ 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 $25 \mathrm{~cm}$
2 $35 \mathrm{~cm}$
3 $30 \mathrm{~cm}$
4 $40 \mathrm{~cm}$
Explanation:
A At balance point - $\frac{5}{\mathrm{R}}=\frac{l_{1}}{100-l_{1}}$ In the second case- $\frac{5}{\mathrm{R} / 2}=\frac{1.6 l_{1}}{100-1.6 l_{1}}$ $\frac{10}{\mathrm{R}}=\frac{1.6 l_{1}}{100-1.6 l_{1}}$ Dividing equation (i) by (ii), $\frac{5}{10}=\frac{l_{1}}{100-l_{1}} \times \frac{100-1.6 l_{1}}{1.6 l_{1}}$ $\frac{1}{2}=\frac{100-1.6 l_{1}}{\left(100-l_{1}\right) \times 1.6}$ $200-3.2 l_{1}=160-1.6 l_{1}$ $1.6 l_{1}=40$ $\therefore \quad l_{1}=\frac{40}{1.6}=25 \mathrm{~cm}$
MHT-CET 2020
Current Electricity
152855
The given circuit is balanced Wheatstone's bridge. The value of resistance ' $x$ ' is
1 $12 \Omega$
2 $4 \Omega$
3 $6 \Omega$
4 $24 \Omega$
Explanation:
A Point B \& D is at same potential At Wheatstone's bridge $\frac{8}{4+x}=\frac{1}{2}$ $16=4+x$ $x=12 \Omega$
MHT-CET 2020
Current Electricity
152856
A galvanometer of resistance $20 \Omega$ has a current sensitivity of $5 \mathrm{div} / \mathrm{mA}$. The instrument has 50 divisions. It can be converted into a voltmeter reading upto 25 volt by connecting a resistance of
1 $2480 \Omega$ in series
2 $20 \Omega$ in parallel
3 $1240 \Omega$ in series
4 $2480 \Omega$ in parallel
Explanation:
A Given that, $\mathrm{G}=20 \Omega$ $\mathrm{V}=25 \mathrm{~V}$ Current sensitivity $=5 \mathrm{div} / \mathrm{mA}$ $\therefore$ For 50 division, the current $\mathrm{I}_{\mathrm{g}}=10 \mathrm{~mA}=10 \times 10^{-3} \mathrm{~A}$ To convert the galvanometer into a voltmeter, a high resistance must be connected in series with the galvanometer coil. This series multiplies $\mathrm{R}_{\mathrm{S}}=\frac{\mathrm{V}}{\mathrm{I}_{\mathrm{g}}}-\mathrm{G}=\frac{25}{10 \times 10^{-3}}-20=2500-20$ $=2480 \Omega \text { in series }$
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
D The given figure is equivalent to- For balanced bridge, $\frac{\mathrm{P}}{\mathrm{R}}=\frac{\mathrm{Q}}{\frac{\mathrm{S}_{1} \mathrm{~S}_{2}}{\mathrm{~S}_{1}+\mathrm{S}_{2}}}$ $\frac{\mathrm{P}}{\mathrm{R}}=\frac{\mathrm{Q}\left(\mathrm{S}_{1}+\mathrm{S}_{2}\right)}{\mathrm{S}_{1} \mathrm{~S}_{2}}$ $\frac{\mathrm{P}}{\mathrm{Q}}=\frac{\mathrm{R}\left(\mathrm{S}_{1}+\mathrm{S}_{2}\right)}{\mathrm{S}_{1} \mathrm{~S}_{2}}$
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 : $-\mathrm{E}=\frac{\mathrm{V}}{\mathrm{L}} \mathrm{x}$ Case II :- Driving cell V remains constant $\mathrm{E}=\frac{\mathrm{V}}{\mathrm{L}^{\prime}} \mathrm{y}$ From equation (i), (ii), we get - $\frac{\mathrm{V}}{\mathrm{L}} \mathrm{x}=\frac{\mathrm{V}}{\mathrm{L}^{\prime}} \mathrm{y}$ $\mathrm{y}=\frac{\mathrm{L}^{\prime}}{\mathrm{L}} \mathrm{x}$ $\because \quad \mathrm{L}^{\prime}>\mathrm{L}$ $\therefore \quad \mathrm{y}>\mathrm{x}$ Hence, balancing length will increase.
152854
In balanced metre bridge $5 \Omega$ in the left gap and $R \Omega$ in the right gap. When $R \Omega$ 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 $25 \mathrm{~cm}$
2 $35 \mathrm{~cm}$
3 $30 \mathrm{~cm}$
4 $40 \mathrm{~cm}$
Explanation:
A At balance point - $\frac{5}{\mathrm{R}}=\frac{l_{1}}{100-l_{1}}$ In the second case- $\frac{5}{\mathrm{R} / 2}=\frac{1.6 l_{1}}{100-1.6 l_{1}}$ $\frac{10}{\mathrm{R}}=\frac{1.6 l_{1}}{100-1.6 l_{1}}$ Dividing equation (i) by (ii), $\frac{5}{10}=\frac{l_{1}}{100-l_{1}} \times \frac{100-1.6 l_{1}}{1.6 l_{1}}$ $\frac{1}{2}=\frac{100-1.6 l_{1}}{\left(100-l_{1}\right) \times 1.6}$ $200-3.2 l_{1}=160-1.6 l_{1}$ $1.6 l_{1}=40$ $\therefore \quad l_{1}=\frac{40}{1.6}=25 \mathrm{~cm}$
MHT-CET 2020
Current Electricity
152855
The given circuit is balanced Wheatstone's bridge. The value of resistance ' $x$ ' is
1 $12 \Omega$
2 $4 \Omega$
3 $6 \Omega$
4 $24 \Omega$
Explanation:
A Point B \& D is at same potential At Wheatstone's bridge $\frac{8}{4+x}=\frac{1}{2}$ $16=4+x$ $x=12 \Omega$
MHT-CET 2020
Current Electricity
152856
A galvanometer of resistance $20 \Omega$ has a current sensitivity of $5 \mathrm{div} / \mathrm{mA}$. The instrument has 50 divisions. It can be converted into a voltmeter reading upto 25 volt by connecting a resistance of
1 $2480 \Omega$ in series
2 $20 \Omega$ in parallel
3 $1240 \Omega$ in series
4 $2480 \Omega$ in parallel
Explanation:
A Given that, $\mathrm{G}=20 \Omega$ $\mathrm{V}=25 \mathrm{~V}$ Current sensitivity $=5 \mathrm{div} / \mathrm{mA}$ $\therefore$ For 50 division, the current $\mathrm{I}_{\mathrm{g}}=10 \mathrm{~mA}=10 \times 10^{-3} \mathrm{~A}$ To convert the galvanometer into a voltmeter, a high resistance must be connected in series with the galvanometer coil. This series multiplies $\mathrm{R}_{\mathrm{S}}=\frac{\mathrm{V}}{\mathrm{I}_{\mathrm{g}}}-\mathrm{G}=\frac{25}{10 \times 10^{-3}}-20=2500-20$ $=2480 \Omega \text { in series }$
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
D The given figure is equivalent to- For balanced bridge, $\frac{\mathrm{P}}{\mathrm{R}}=\frac{\mathrm{Q}}{\frac{\mathrm{S}_{1} \mathrm{~S}_{2}}{\mathrm{~S}_{1}+\mathrm{S}_{2}}}$ $\frac{\mathrm{P}}{\mathrm{R}}=\frac{\mathrm{Q}\left(\mathrm{S}_{1}+\mathrm{S}_{2}\right)}{\mathrm{S}_{1} \mathrm{~S}_{2}}$ $\frac{\mathrm{P}}{\mathrm{Q}}=\frac{\mathrm{R}\left(\mathrm{S}_{1}+\mathrm{S}_{2}\right)}{\mathrm{S}_{1} \mathrm{~S}_{2}}$
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 : $-\mathrm{E}=\frac{\mathrm{V}}{\mathrm{L}} \mathrm{x}$ Case II :- Driving cell V remains constant $\mathrm{E}=\frac{\mathrm{V}}{\mathrm{L}^{\prime}} \mathrm{y}$ From equation (i), (ii), we get - $\frac{\mathrm{V}}{\mathrm{L}} \mathrm{x}=\frac{\mathrm{V}}{\mathrm{L}^{\prime}} \mathrm{y}$ $\mathrm{y}=\frac{\mathrm{L}^{\prime}}{\mathrm{L}} \mathrm{x}$ $\because \quad \mathrm{L}^{\prime}>\mathrm{L}$ $\therefore \quad \mathrm{y}>\mathrm{x}$ Hence, balancing length will increase.
152854
In balanced metre bridge $5 \Omega$ in the left gap and $R \Omega$ in the right gap. When $R \Omega$ 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 $25 \mathrm{~cm}$
2 $35 \mathrm{~cm}$
3 $30 \mathrm{~cm}$
4 $40 \mathrm{~cm}$
Explanation:
A At balance point - $\frac{5}{\mathrm{R}}=\frac{l_{1}}{100-l_{1}}$ In the second case- $\frac{5}{\mathrm{R} / 2}=\frac{1.6 l_{1}}{100-1.6 l_{1}}$ $\frac{10}{\mathrm{R}}=\frac{1.6 l_{1}}{100-1.6 l_{1}}$ Dividing equation (i) by (ii), $\frac{5}{10}=\frac{l_{1}}{100-l_{1}} \times \frac{100-1.6 l_{1}}{1.6 l_{1}}$ $\frac{1}{2}=\frac{100-1.6 l_{1}}{\left(100-l_{1}\right) \times 1.6}$ $200-3.2 l_{1}=160-1.6 l_{1}$ $1.6 l_{1}=40$ $\therefore \quad l_{1}=\frac{40}{1.6}=25 \mathrm{~cm}$
MHT-CET 2020
Current Electricity
152855
The given circuit is balanced Wheatstone's bridge. The value of resistance ' $x$ ' is
1 $12 \Omega$
2 $4 \Omega$
3 $6 \Omega$
4 $24 \Omega$
Explanation:
A Point B \& D is at same potential At Wheatstone's bridge $\frac{8}{4+x}=\frac{1}{2}$ $16=4+x$ $x=12 \Omega$
MHT-CET 2020
Current Electricity
152856
A galvanometer of resistance $20 \Omega$ has a current sensitivity of $5 \mathrm{div} / \mathrm{mA}$. The instrument has 50 divisions. It can be converted into a voltmeter reading upto 25 volt by connecting a resistance of
1 $2480 \Omega$ in series
2 $20 \Omega$ in parallel
3 $1240 \Omega$ in series
4 $2480 \Omega$ in parallel
Explanation:
A Given that, $\mathrm{G}=20 \Omega$ $\mathrm{V}=25 \mathrm{~V}$ Current sensitivity $=5 \mathrm{div} / \mathrm{mA}$ $\therefore$ For 50 division, the current $\mathrm{I}_{\mathrm{g}}=10 \mathrm{~mA}=10 \times 10^{-3} \mathrm{~A}$ To convert the galvanometer into a voltmeter, a high resistance must be connected in series with the galvanometer coil. This series multiplies $\mathrm{R}_{\mathrm{S}}=\frac{\mathrm{V}}{\mathrm{I}_{\mathrm{g}}}-\mathrm{G}=\frac{25}{10 \times 10^{-3}}-20=2500-20$ $=2480 \Omega \text { in series }$
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
D The given figure is equivalent to- For balanced bridge, $\frac{\mathrm{P}}{\mathrm{R}}=\frac{\mathrm{Q}}{\frac{\mathrm{S}_{1} \mathrm{~S}_{2}}{\mathrm{~S}_{1}+\mathrm{S}_{2}}}$ $\frac{\mathrm{P}}{\mathrm{R}}=\frac{\mathrm{Q}\left(\mathrm{S}_{1}+\mathrm{S}_{2}\right)}{\mathrm{S}_{1} \mathrm{~S}_{2}}$ $\frac{\mathrm{P}}{\mathrm{Q}}=\frac{\mathrm{R}\left(\mathrm{S}_{1}+\mathrm{S}_{2}\right)}{\mathrm{S}_{1} \mathrm{~S}_{2}}$
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 : $-\mathrm{E}=\frac{\mathrm{V}}{\mathrm{L}} \mathrm{x}$ Case II :- Driving cell V remains constant $\mathrm{E}=\frac{\mathrm{V}}{\mathrm{L}^{\prime}} \mathrm{y}$ From equation (i), (ii), we get - $\frac{\mathrm{V}}{\mathrm{L}} \mathrm{x}=\frac{\mathrm{V}}{\mathrm{L}^{\prime}} \mathrm{y}$ $\mathrm{y}=\frac{\mathrm{L}^{\prime}}{\mathrm{L}} \mathrm{x}$ $\because \quad \mathrm{L}^{\prime}>\mathrm{L}$ $\therefore \quad \mathrm{y}>\mathrm{x}$ Hence, balancing length will increase.
