Answer
Verified
91.5k+ views
Hint At first, calculate the voltage for the resistance $R$ and then, calculate the voltage reading for the voltmeter of equivalent resistance. Then, by using the expressions for voltage for $R$ and equivalent resistance for voltmeter for the 95% accuracy.
Complete Step by Step Solution
In the above figure, we have to use the Ohm’s law which states that the current through the conductor between two points is directly proportional to the voltage across the two points. So, by using the ohm’s law we can find the value for ${V_0}$ as –
${V_0} = ir \cdots \left( 1 \right)$
Now, drawing the figure with the voltmeter of resistance ${R_v}$ -
Now, finding the equivalent resistance in the above figure –
From the above figure, we can see that the resistances $R$ and ${R_v}$ are parallel to each other. So, -
$
\Rightarrow \dfrac{1}{{{R_{eq}}}} = \dfrac{1}{R} + \dfrac{1}{{{R_v}}} \\
\Rightarrow {R_{eq}} = \dfrac{{R{R_v}}}{{R + {R_v}}} \\
$
Again, using the ohm’s law, finding the expression for voltage –
$V = i\dfrac{{R{R_v}}}{{R + {R_v}}} \cdots \left( 2 \right)$
Now, according to the question, it is given that, accuracy should be 95% then, we get –
$ \Rightarrow V = 0.95{V_0}$
Using the value of $V$ and ${V_0}$ from equation (1) and equation (2), we get –
$ \Rightarrow i\dfrac{{R{R_v}}}{{R + {R_v}}} = 0.95\left( {iR} \right)$
By further solving, we get –
$
\Rightarrow {R_v} > 0.95R + 0.95{R_v} \\
\Rightarrow 0.05{R_v} > 0.95R \\
\Rightarrow {R_v} > \dfrac{{0.95}}{{0.05}}R \\
\therefore {R_v} > 19R \\
$
Hence, from the above expression we can see that option (D) is the correct answer.
Note As the accuracy should be 95% so, it can be rewritten as $\dfrac{{95}}{{100}}$ after solving this we get the efficiency as 0.95 which is to be used during the solving of the problem.
Complete Step by Step Solution
In the above figure, we have to use the Ohm’s law which states that the current through the conductor between two points is directly proportional to the voltage across the two points. So, by using the ohm’s law we can find the value for ${V_0}$ as –
${V_0} = ir \cdots \left( 1 \right)$
Now, drawing the figure with the voltmeter of resistance ${R_v}$ -
Now, finding the equivalent resistance in the above figure –
From the above figure, we can see that the resistances $R$ and ${R_v}$ are parallel to each other. So, -
$
\Rightarrow \dfrac{1}{{{R_{eq}}}} = \dfrac{1}{R} + \dfrac{1}{{{R_v}}} \\
\Rightarrow {R_{eq}} = \dfrac{{R{R_v}}}{{R + {R_v}}} \\
$
Again, using the ohm’s law, finding the expression for voltage –
$V = i\dfrac{{R{R_v}}}{{R + {R_v}}} \cdots \left( 2 \right)$
Now, according to the question, it is given that, accuracy should be 95% then, we get –
$ \Rightarrow V = 0.95{V_0}$
Using the value of $V$ and ${V_0}$ from equation (1) and equation (2), we get –
$ \Rightarrow i\dfrac{{R{R_v}}}{{R + {R_v}}} = 0.95\left( {iR} \right)$
By further solving, we get –
$
\Rightarrow {R_v} > 0.95R + 0.95{R_v} \\
\Rightarrow 0.05{R_v} > 0.95R \\
\Rightarrow {R_v} > \dfrac{{0.95}}{{0.05}}R \\
\therefore {R_v} > 19R \\
$
Hence, from the above expression we can see that option (D) is the correct answer.
Note As the accuracy should be 95% so, it can be rewritten as $\dfrac{{95}}{{100}}$ after solving this we get the efficiency as 0.95 which is to be used during the solving of the problem.
Recently Updated Pages
Name the scale on which the destructive energy of an class 11 physics JEE_Main
Write an article on the need and importance of sports class 10 english JEE_Main
Choose the exact meaning of the given idiomphrase The class 9 english JEE_Main
Choose the one which best expresses the meaning of class 9 english JEE_Main
What does a hydrometer consist of A A cylindrical stem class 9 physics JEE_Main
A motorcyclist of mass m is to negotiate a curve of class 9 physics JEE_Main
Other Pages
Electric field due to uniformly charged sphere class 12 physics JEE_Main
If a wire of resistance R is stretched to double of class 12 physics JEE_Main
Derive an expression for maximum speed of a car on class 11 physics JEE_Main
3 mole of gas X and 2 moles of gas Y enters from the class 11 physics JEE_Main
A soldier with a machine gun falling from an airplane class 11 physics JEE_MAIN
Velocity of car at t 0 is u moves with a constant acceleration class 11 physics JEE_Main