A battery is connected to a potentiometer and a balance point is obtained at 84 cm along the wire. When its terminals are connected by a $5 \Omega$ resistor, the balance point changes to 70 cm. Find the new position of the balance point when $5 \Omega$ resistor is changed by $4 \Omega$ resistor.
A. 26.5 cm
B. 52 cm
C. 67.2 cm
D. 83.3 cm
Answer
619.5k+ views
Hint: To solve this problem, use the formula for internal resistance. Substitute the values in the formula of internal resistance for $5 \Omega$ resistor and calculate the internal resistance for it. Then, use the same formula to find the internal resistance for $4 \Omega$ resistor. Substitute the internal resistance obtained above in this equation and find the unknown variable which is the new position of the balance point.
Formula used:
$r= \dfrac {{l}_{1}-{l}_{2}}{{l}_{2}}R$
Complete answer:
Given: ${l}_{1}= 84 cm$
${l}_{2}= 70 cm$
${R}_{1}= 5 \Omega$
${R}_{2}= 4 \Omega$
The formula for internal resistance is given by,
$r= \dfrac {{l}_{1}-{l}_{2}}{{l}_{2}}R$
Internal resistance when $5 \Omega$ resistor is connected is given by,
$r= \dfrac {{l}_{1}-{l}_{2}}{{l}_{2}}{R}_{1}$
Substituting values in above equation we get,
$r= \dfrac {84-70}{70}\times 5$
$\Rightarrow r= \dfrac {14}{70}\times 5$
$\Rightarrow r= 1 \Omega$
Internal resistance when $4 \Omega$ resistor is connected is given by,
$r= \dfrac {{l}_{1}-{l}_{3}}{{l}_{3}}{R}_{2}$
Substituting values in above equation we get,
$1= \dfrac {84-{l}_{3}}{{l}_{3}} \times 4$
$\Rightarrow {l}_{3}=(84-{l}_{3}) \times 4$
$\Rightarrow {l}_{3}= 336 – 4{l}_{3}$
$\Rightarrow 5{l}_{3}= 336$
$\Rightarrow {l}_{3}= \dfrac {336}{5}$
$\Rightarrow {l}_{3}= 67.2 cm$
Hence, when $5 \Omega$ resistor is changed by $4 \Omega$ resistor, the new position of the balance point is 67.2 cm.
So, the correct answer is option C i.e. 67.2 cm.
Note:
Internal resistance is the opposition offered by the cells and batteries to the flow of current flowing in the generation of heat. Internal resistance depends upon the nature of the material of the wire. The potentiometer is an arrangement that can be used to find the unknown value of resistances and the cell’s internal resistance. Potentiometer is also used to determine the unknown values of potential differences.
Formula used:
$r= \dfrac {{l}_{1}-{l}_{2}}{{l}_{2}}R$
Complete answer:
Given: ${l}_{1}= 84 cm$
${l}_{2}= 70 cm$
${R}_{1}= 5 \Omega$
${R}_{2}= 4 \Omega$
The formula for internal resistance is given by,
$r= \dfrac {{l}_{1}-{l}_{2}}{{l}_{2}}R$
Internal resistance when $5 \Omega$ resistor is connected is given by,
$r= \dfrac {{l}_{1}-{l}_{2}}{{l}_{2}}{R}_{1}$
Substituting values in above equation we get,
$r= \dfrac {84-70}{70}\times 5$
$\Rightarrow r= \dfrac {14}{70}\times 5$
$\Rightarrow r= 1 \Omega$
Internal resistance when $4 \Omega$ resistor is connected is given by,
$r= \dfrac {{l}_{1}-{l}_{3}}{{l}_{3}}{R}_{2}$
Substituting values in above equation we get,
$1= \dfrac {84-{l}_{3}}{{l}_{3}} \times 4$
$\Rightarrow {l}_{3}=(84-{l}_{3}) \times 4$
$\Rightarrow {l}_{3}= 336 – 4{l}_{3}$
$\Rightarrow 5{l}_{3}= 336$
$\Rightarrow {l}_{3}= \dfrac {336}{5}$
$\Rightarrow {l}_{3}= 67.2 cm$
Hence, when $5 \Omega$ resistor is changed by $4 \Omega$ resistor, the new position of the balance point is 67.2 cm.
So, the correct answer is option C i.e. 67.2 cm.
Note:
Internal resistance is the opposition offered by the cells and batteries to the flow of current flowing in the generation of heat. Internal resistance depends upon the nature of the material of the wire. The potentiometer is an arrangement that can be used to find the unknown value of resistances and the cell’s internal resistance. Potentiometer is also used to determine the unknown values of potential differences.
Recently Updated Pages
Difference Between Prokaryotic Cells and Eukaryotic Cells

If x a + bt + ct2 where x is in meters and t is in class 11 physics CBSE

A car covers the first half distance between two places class 11 physics CBSE

The resultant of two vectors overrightarrow P and overrightarrow class 11 physics CBSE

Find the value of cos 135 class 11 maths CBSE

A mass M is held in place by an applied force F and class 11 physics CBSE

Trending doubts
One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

Find the value of the expression given below sin 30circ class 11 maths CBSE

Draw a diagram of nephron and explain its structur class 11 biology CBSE

10 examples of friction in our daily life

Proton was discovered by A Thomson B Rutherford C Chadwick class 11 chemistry CBSE

Bond order ofO2 O2+ O2 and O22 is in order A O2 langle class 11 chemistry CBSE

