NEET Physics Questions: Experimental Skills

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N divisions on the main scale of a vernier callipers coincides with (N+1) divisions on the vernier scale. If each division on the mainscale is of a units, the least of count of instrument is




(N+1) divisions on the vernier scale = N divisions on main scale $ \therefore $ 1 division on vernier scale = N/N + 1 divisions on main scale Each division on the main scale in of a units $ \therefore $ 1 division on vernier scale (N /N+1 ) a units = a' (say) Least count = 1 main scale division -1 vernier scale division = a - a' = a -(N / N+1 ) a = a/N+1
The edge of a cube is measured using a vernier caliper (9 divisions of the main scale is equal to 10 divisions of vernier scale and 1 main scale division is 1mm). The main scale division reading is 10 and 1 division of vernier scale was found to be coinciding with the mainscale. The mass of the cube is 2.736 g. What will be the density in $g/cm^3 $ upto correct significant figures ?




1 MSD = 1 mm 9 MSD = 10 VSD $ \therefore $ Least count, LC = 1MSO - 1VSO = 1mm - 9/10 mm = 1/10 mm Measure reading of edge =MSR + VSR (LC) = 10 + 1 1 /10 = 10.1 mm $ Volume of cube V = (101)^3 cm^3 = 1.03 cm^3$ After rounding off upto 3 significant digits, as edge length is measured upto 3 significant digits] Density of cube = 2736 / 1.03 = 2.6563 g/cm^3 = 2.66 g/cm^3 (After rounding offto 3 significant digits)
When the edge of the circular scale lies to the left of O mark on the main scale, when the stud and spindle touch each other, Then what will be the zero error ?




When the edge of the circular scale lies to the left of the O mark on the main scale and the stud and spindle touch each other, the zero error is negative. This happens because the zero of the vernier scale is to the left of the zero of the main scale.
When the edge of the circular scale lies to the right of the O mark on the main scale, when the stud and the spindle touch each other,then what will be the zero error ?




Here zero error is positive, so zero correction is negative.
When the screw and stud touch each other, the edges of a certain screw gauge is on left of the O mark on the main scale and the 96th division of the circular scale coincides with the circular line of graduation then what is the value of zero error ?




Here the zero error is negative. $ Z = (96 -100) \times LC = -4 \times 0.01 mm = - 0.04 mm$
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When the screw and stud touch each other, the edge of a certain screw gauge is to the right of the O mark on the main scale and 5th division of the circular scale coincides with the line of graduation, then what is the value of zero error ?




Here the zero error is positive $ z = 5 \times L.C = 5 \times 0.01 mm = + 0.05 mm$
Screw guage A has a pitch of 1 mm and 50 division on its circular scale screw guage B has a pitch of 0.5 mm and 100 divisions on its circular scale. If (LxC) is least count, then which possibility is true ? What is the parilrility ?




Least count = Pithc / No. of divisions on circular scale
The pitch of screw gauge is 1 mm and there are 100 divisions on the circular scale. What measuring the diameter of a wire, the linear scale reads, 1 mm and 47th division on the circular scale coincides with the reference line. The length of the wire is 5.6 cm. What will be the curved surface area (in cm2) of the wire in appropriate number of significant figures.




$ Curved surface area = 2 \pi rl $
Complete the following sentance. Time period ofoscillatin of a simple pendulumis dependent on............




$ T \alpha \sqrt l $
Complete the following seutence In a damped oscillation of a pendulam .........




knowledge base
If the time period of undamped oscillation is T and that of damped oscillatin is $T^1$ ,then what is the relation between T &




$ T \alpha { 1 \over \omega } \omega = \sqrt { k \over m } \Rightarrow 2 \pi \sqrt { m \over k } $ $ where as T' \alpha { 1 \over \omega '} where \omega' = \sqrt { {k\over m} = { b^2 \over 4m^2} } $ $as \omega' \lt \omega \Rightarrow T' \gt T $
The energy dissiplated in a damped oscillation ...........




$ E = E_0 e ^ { - \lambda t } $
What is the equatin for a damped oscillator, where k and b are constants and x is displacement.




$ F = - kx -by \Rightarrow m { d^2 x \over dt^2 } + { bdx \over dt } + kx =0$
In a damped oscillatin with damping constant b. The time takenfor amplitude of oscillatin to drop to half what is its initial value ?




$ A = A_0 e ^ {-bt /m} , t = T _{1/2 } whrn A = { A_0 \over 2 } $
In a damped oscillatin with damping constant b. The time takenfor its mechanical energy to drop to half. What is its value ?




$ E =E_0 e ^ { -bt / m } when E = { E_0 \over 2 } , t = T _ {1/2} $