Q. No.If force \([F]\), acceleration \([A]\) and time \([T]\) are chosen as the fundamental physical quantities, then find the dimensions of energy:
1.
\(\left[FAT^{-1}\right]\)
2.
\(\left[FA^{-1}T\right]\)
3.
\(\left[FAT\right]\)
4.
\(\left[FAT^{2}\right]\)
1. \([ML^0T^0]\)
2. \([M^2L^{-2}T^{-1}]\)
3. \([M^2L^{-1}T^{0}]\)
4. \([ML^{-1}T^{-1}]\)
Time intervals measured by a clock give the following readings:
\(1.25~\text{s},~1.24~\text{s}, ~1.27~\text{s},~1.21~\text{s},~1.28~\text{s}.\)
What is the percentage relative error of the observations?
1. \(2\)%
2. \(4\)%
3. \(16\)%
4. \(1.6\)%
1. \(2.91 \times 10^{-4}~\text{rad} \)
2. \(4.85 \times 10^{-4}~\text{rad} \)
3. \(4.80 \times 10^{-6} ~\text{rad} \)
4. \(1.75 \times 10^{-2}~\text{rad} \)
| 1. | both units and dimensions |
| 2. | units but no dimensions |
| 3. | dimensions but no units |
| 4. | no units and no dimensions |
1. electric permittivity
2. magnetic flux
3. self-inductance
4. magnetic permeability
1. \(14\times10^{2}\)
2. \(138\times10^{1}\)
3. \(1382\)
4. \(1382.5\)
When the circular scale of a screw gauge completes \(2\) rotations, it covers \(1\) mm over the pitch scale. The total number of circular scale divisions is \(50.\) The least count of the screw gauge in metres is:
1. \(10^{-4}\)
2. \(10^{-5}\)
3. \(10^{-2}\)
4. \(10^{-3}\)
The determination of the value of acceleration due to gravity \((g)\) by simple pendulum method employs the formula,
\(g=4\pi^2\dfrac{L}{T^2}\)
The expression for the relative error in the value of \(g\) is:
| 1. | \(\dfrac{\Delta g}{g}=\dfrac{\Delta L}{L}+2\Big(\dfrac{\Delta T}{T}\Big)\) |
| 2. | \(\dfrac{\Delta g}{g}=4\pi^2\Big[\dfrac{\Delta L}{L}-2\dfrac{\Delta T}{T}\Big]\) |
| 3. | \(\dfrac{\Delta g}{g}=4\pi^2\Big[\dfrac{\Delta L}{L}+2\dfrac{\Delta T}{T}\Big]\) |
| 4. | \(\dfrac{\Delta g}{g}=\dfrac{\Delta L}{L}-2\Big(\dfrac{\Delta T}{T}\Big)\) |
1. Force
2. Momentum
3. Young's modulus of elasticity
4. Coefficient of viscosity
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