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A physical quantity ‘\( y \)’ is represented by the formula \( y = m^{2} r^{-4} g^{x} l^{-\dfrac{3}{2}} \). If the percentage errors found in \( y, m, r, l \) and \( g \) are \( 18 \), \( 1 \), \( 0.5 \), \( 4 \) and \( p \) respectively, then find the value of \( x \) and \( p \).

Assertion \( A \): If in five complete rotations of the circular scale, the distance travelled on main scale of the screw gauge is \( 5 \, \text{mm} \) and there are \( 50 \) total divisions on circular scale, then least count is \( 0.001 \, \text{cm} \).
Reason \( R \): Least Count \( = \dfrac{\text{Pitch}}{\text{Total divisions on circular scale}} \).
In the light of the above statements, choose the most appropriate answer from the options given below:

Student A and Student B used two screw gauges of equal pitch and \( 100 \) equal circular divisions to measure the radius of a given wire. The actual value of the radius of the wire is \( 0.322 \, \text{cm} \). The absolute value of the difference between the final circular scale readings observed by the students A and B is __________________ . [Figure shows position of reference ‘\( O \)’ when jaws of screw gauge are closed]
Given pitch = \( 0.1 \, \text{cm} \).

In a Screw Gauge, fifth division of the circular scale coincides with the reference line when the ratchet is closed. There are \( 50\) divisions on the circular scale, and the main scale moves by \( 0.5 \text{mm} \) on a complete rotation. For a particular observation the reading on the main scale is \( 5 \text{mm} \) and the \( 20^{\text{th}} \) division of the circular scale coincides with reference line. Calculate the true reading.

A student determined Young’s Modules of elasticity using the formula \( Y=\dfrac{MgL^{3}}{4bd^{3}\,\delta} \). The value of \( g \) is taken to be \( 9.8 \, \text{m} / \text{s}^{2} \), without any significant error, his observation are as following.

Then the fractional error in the measurement of \( Y \) is:

Two resistor \( R_{1}=(4 \pm 0.8)\, \Omega \) and \( R_{2}=(4 \pm 0.4)\, \Omega \) are connected in parallel. The equivalent resistance of their parallel combination will be:

The acceleration due to gravity is found upto an accuracy of \( 4\% \) on a planet. The energy supplied to a simple pendulum of known mass \( m \) to undertake oscillations of time period \( T \) is being estimated. If time period is measured to an accuracy of \( 3\% \), the accuracy to which \( E \) is known as __________ \( \% \).

The time period of a simple pendulum is given by \( T = 2\pi \sqrt{\frac{l}{g}} \). The measured value of the length of pendulum is \( 10 \, \text{cm} \) known to a \( 1 \, \text{mm} \) accuracy. The time for \( 200 \) oscillations of the pendulum is found to be \( 100 \, \text{second} \) using a clock of \( 1 \, \text{s} \) resolution. The percentage accuracy in the determination of \( g \) using this pendulum is \( x \). The value of \( x \) to the nearest integer is:

If the length of the pendulum in pendulum clock increases by \( 0.1\% \), then the error in time per day is:

In the experiment of Ohm’s law, a potential difference of \( 5.0 \, \text{V} \) is applied across the end of a conductor of length \( 10.0 \, \text{cm} \) and diameter of \( 5.00 \, \text{mm} \). The measured current in the conductor is \( 2.00 \, \text{A} \). The maximum permissible percentage error in the resistivity of the conductor is:

The radius of a sphere is measured to be \( (7.50 \pm 0.85)\, \text{cm} \). Suppose the percentage error in its volume is \( x \). The value of \( x \), to the nearest \( x \), is:

The diameter of a spherical bob is measured using a vernier callipers. \( 9 \) divisions of the main scale, in the vernier callipers, are equal to \( 10 \) divisions of vernier scale. One main scale division is \( 1 \, \text{mm} \). The main scale reading is \( 10 \, \text{mm} \) and \( 8^{\text{th}} \) division of vernier scale was found to coincide exactly with one of the main scale division. If the given vernier calliper has positive zero error of \( 0.04 \, \text{cm} \), then the radius of the bob is ______ \( \times 10^{-2} \, \text{cm} \) .

The period of oscillation of a simple pendulum is \( T = 2\pi\sqrt{\frac{L}{g}} \). Measured value of ‘L’ is \( 1.0 \, \text{m} \) from meter scale having a minimum division of \( 1 \, \text{mm} \) and time of one complete oscillation is \( 1.95 \, \text{s} \) measured from stopwatch of \( 0.01 \, \text{s} \) resolution. The percentage error in the determination of ‘g’ will be:

The vernier scale used for measurement has a positive zero error of \( 0.2 \, \text{mm} \). If while taking a measurement it was noted that ‘\( 0 \)’ on the vernier scale lies between \( 8.5 \, \text{cm} \) and \( 8.6 \, \text{cm} \), vernier coincidence is \( 6 \), then the correct value of measurement is ______ \( \text{cm} \). (least count = \( 0.01 \, \text{cm} \))

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JEE Main 2021 26th August Evening Shift

If the length of the pendulum in pendulum clock increases by \( 0.1\% \), then the error in time per day is: