Suppose the kinetic energy of a body oscillating with amplitude A and at a distance x is given by \(K=\frac{Bx}{x^{2}+A^{2}}\) The dimensions of Bare the same as that of

work/time
work × distance
work/distance
work × time

Solution

From K=\(\frac{Bx}{x^{2}+A^{2}}=\frac{Bx}{x^{2}}=\frac{B}{x}\)

∴ B = K × x = K.E. × distance= work × distance

The dimensions of voltage in terms of mass (M), length (L)and time (T) and ampere (A) are

[M L² T^-2 A^-2]
[M L² T³ A^-1]
[M L² T^-3 A¹]
[M L² T^-3 A^-1]

Solution

[V]=\(\left [ \frac{W}{Q} \right ]=\frac{ML^{2}T^{-2}}{AT}=M\, L^{2}\, A^{-1}\, T^{-3}\)

The frequency of vibration of a string is given by f =\(\frac{n}{2L}\sqrt{\frac{T}{m}}\), where T is tension in the string, L is the length, n is number of harmonics. The dimensional formula for m is

[M⁰ L T]
[M¹ L^-1 T^-1]
[M¹ L^-1 T⁰]
[M⁰ L T^-1]

Solution

Clearly, m=\(\frac{n^{2}T}{4f^{2}L^{2}};[m]=\frac{MLT^{-2}}{T^{-2}L^{2}}\)

A force is given by F = at + bt², where t is time, the dimensions of a and b are

[M L T^–4] and [M L T^-1]
[M L T^-1] and [M L T⁰]
[M L T^–3] and [M L T^–4]
[M L T^–3] and [M L T⁰]

Solution

[at] = [F] amd [bt²] = [F]

⇒[a] = MLT^–3 and [b] = M L T^–4

In the equation X = 3YZ², X and Z are dimensions of capacitance and magnetic induction respectively. In MKSQ system, the dimensional formula for Y is

[M^-3 L^-2 T^-2 Q^-4]
[M L^-2]
[M^-3 L^-2 Q⁴ T⁸]
[M^-3 L^-2 Q⁴ T⁴]

Solution

[Y]=\(\frac{[X]}{[Z^{2}]}=\frac{M^{-1}L^{-2}T^{4}A^{2}}{M^{2}T^{-4}A^{-2}}\)=M^-3L^-2Q⁴T⁴

\(\left ( A=\frac{Q}{T} \right )\)

The equation of a wave is given by y=a sin ω\(\left ( \frac{X}{V}-K \right )\), where ω is angular velocity and v is linear velocity. The dimensions of K will be

[T²]
[T^-1]
[T]
[LT]

Solution

The quantity \(\left ( \frac{\omega x}{v}-\omega k \right )\) has dimension of angle and hence ωk is dimensionless being angle.

Which one of the following has the same dimension as that of time, if R is resistance, L inductance and C is capacitance?

RC
\(\sqrt{LC}\)
L/R
All of the above

Solution

\([\sqrt{LC}]=\sqrt{ML^{2}T^{-2}A^{-2}.M^{-1}L^{-2}T^{4}A^{2}}=T\)

Consider the following pairs of quantities

1.Young’s modulus; pressure
2.Torque; energy
3.Linear momentum; work
4.Solar day; light year.
In which cases are the dimensions, within a pair, same?

1 and 3
1 and 4
1 and 2
2 and 4

The dimensions of a rectangular block measured with callipers having least count of 0.01 cm are 5 mm × 10 mm × 5mm. The maximum percentage error in the measurement of the volume of the block is

Solution

% error =\(\frac{0.01}{0.5}\times 100+\frac{0.01}{1.0}\times 100+\frac{0.01}{0.5}\times 100\)

= 2 + 1+ 2= 4 + 1 = 5

In a simple pendulum experiment for the determination of acceleration due to gravity, time period is measured with an accuracy of 0.2% while length was measured with an accuracy of 0.5%. The percentage accuracy in the value of g so obtained is

0.25%
0.7%
0.9%
1.0%

Solution

T=\(2\pi \sqrt{\frac{\iota }{g}}\),g α \(\frac{\iota }{T^{2}}\)

∴ \(\frac{\Delta g}{g}\times 100\)= 0.5% +2 × 0.2% = 0.9%

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