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NEET • PHYSICS • UNITS & MEASUREMENTS

Units & measurements practice · Hard

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31 Hard
31 of 41
Application +4 | -1
In gas equation (P+aV2)(V−b)=RT\left( {P + \frac{a}{{{V^2}}}} \right)\left( {V - b} \right) = RT, where T is absolute temperature, P is the pressure and V is the volume of the gas. The dimensional formula of (aV2)\left( {\frac{a}{{{V^2}}}} \right) is
View answer and explanation
Correct option: C
ML−1T−2M{L^{ - 1}}{T^{ - 2}}
Option C (ML−1T−2M{L^{ - 1}}{T^{ - 2}}) is correct because the term a/V² must have the same dimensions as pressure (P).
32 Hard
32 of 41
Application +4 | -1
The reverberation period T of a room depends on its volume V, its surface area A and velocity of sound C. If K is dimensionless constant then T =
View answer and explanation
Correct option: A
KVCA\frac{{KV}}{{CA}}
Option A (KVCA\frac{{KV}}{{CA}}) is correct because it correctly relates the reverberation period to volume, surface area, and velocity of sound using dimensional analysis.
33 Hard
33 of 41
Application +4 | -1
If the error in the measurement of the volume of a sphere is 4.5%, then error in the measurement of its surface area is
View answer and explanation
Correct option: B
0.03
Option B (0.03) is correct because the error in surface area is twice the error in the radius, and the error in volume is three times the error in the radius.
34 Hard
34 of 41
Application +4 | -1
A gas bubble from an explosion under water oscillates with a time period T, depends upon static pressure p, density of water ρ\rho and the total energy of explosion E. Find the expression for the time period T.(where, k is a dimensionless constant)
View answer and explanation
Correct option: A
T=kp−5/6ρ1/2E1/3T = k{p^{ - 5/6}}{\rho ^{1/2}}{E^{1/3}}
Option A (T=kp−5/6ρ1/2E1/3T = k{p^{ - 5/6}}{\rho ^{1/2}}{E^{1/3}}) is correct because dimensional analysis confirms that the dimensions of this expression match the dimensions of time.
35 Hard
35 of 41
Application +4 | -1
If K denotes coefficient of thermal conductivity, d the density and c the specific heat, the unit of X, where X=K/dc will be
View answer and explanation
Correct option: D
cm2sec⁡−1  \mathrm{c}{\mathrm{m}^2}{\sec ^{ - 1}}\;
Option D (cm2sec⁡−1\mathrm{c}{\mathrm{m}^2}{\sec ^{ - 1}}) is correct because the unit of X, where X=K/dc, is calculated by substituting the units of K, d, and c.
36 Hard
36 of 41
Analyzing +4 | -1
The focal length of a mirror is given by 1f=1u+1  v\frac{1}{f} = \frac{1}{u} + \frac{{1\;}}{v} where u and vv represent object and image distances respectively. The maximum relative error in ff is
View answer and explanation
Correct option: D
Δff=Δuu+Δvv+Δuu+v+Δvu+v\frac{{\Delta f}}{f} = \frac{{\Delta u}}{u} + \frac{{\Delta v}}{v} + \frac{{\Delta u}}{{u + v}} + \frac{{\Delta v}}{{u + v}}
Option D (Δff=Δuu+Δvv+Δuu+v+Δvu+v\frac{{\Delta f}}{f} = \frac{{\Delta u}}{u} + \frac{{\Delta v}}{v} + \frac{{\Delta u}}{{u + v}} + \frac{{\Delta v}}{{u + v}}) is correct because it correctly represents the maximum relative error in f, derived using error propagation techniques.
37 Hard
37 of 41
Analyzing +4 | -1
A student performs an experiment for determination of g(=4π2lT2),  l≈1m,\mathrm{g}\left( { = \frac{{4{\mathrm{\pi }^2}l}}{{{T^2}}}} \right),\mathrm{\;}l \approx 1\mathrm{m},, and he commits an error of Δl.\Delta l.. For T he takes the time of n oscillations with the stop watch of least count ∆T and he commits a human error of 0.1 s. For which of the following data, the measurement of g will be most accurate?
View answer and explanation
Correct option: D
ΔL=0.5,  ΔT=0.05,  n=50\Delta L = 0.5,\;\Delta T = 0.05,\;n = 50
Option D (ΔL=0.5,  ΔT=0.05,  n=50\Delta L = 0.5,\;\Delta T = 0.05,\;n = 50) is correct because it minimizes the percentage error in g by having a smaller error in time measurement and a larger number of oscillations.
38 Hard
38 of 41
Analyzing +4 | -1
If velocity of light c, acceleration due to gravity g and density d are the fundamental quantities in a system of measurement then mass has
View answer and explanation
Correct option: D
dimensions 6 in c
Option D (dimensions 6 in c) is correct because the mass can be expressed as M=c6g−2d−1M = c^6 g^{-2} d^{-1} using dimensional analysis.
39 Hard
39 of 41
Application +4 | -1
E, m, L and G denote energy, mass, angular momentum and gravitational constant respectively, EL2m5G2\frac{{E{L^2}}}{{{m^5}{G^2}}} has the dimensions of
View answer and explanation
Correct option: D
angle
Option D (angle) is correct because it involves finding the dimensions of EL2m5G2\frac{{E{L^2}}}{{{m^5}{G^2}}}, where E is energy, L is angular momentum, m is mass, and G is the gravitational constant. After substituting the dimensions of each quantity and simplifying, the result is dimensionless, which corresponds to an angle.
40 Hard
40 of 41
Application +4 | -1
Taking frequency ff, velocity vv and density ρ\rho to be the fundamental quantities, then the dimensional formula for momentum will be
View answer and explanation
Correct option: A
[ρv4f−3]\left[ {\rho {v^4}{f^{ - 3}}} \right]
Option A ([ρv4f−3\rho {v^4}{f^{ - 3}}]) is correct because dimensional analysis, using frequency, velocity, and density as fundamental quantities, yields the dimensions of momentum.
41 Hard
41 of 41
Application +4 | -1
The external and internal diameters of a hallow cylinder are measured to be (4.23±0.01) cm(4.23 \pm 0.01)\,cm and (3.89±0.01) cm(3.89 \pm 0.01)\,cm. The thickness of the wall of the cylinder is
View answer and explanation
Correct option: C
(0.17±0.01) cm(0.17 \pm 0.01)\,cm
Option C ((0.17±0.01) cm(0.17 \pm 0.01)\,cm) is correct because it accurately calculates the thickness and its associated error.
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