A screw gauge has the least count of \(0.01~\text{mm}\) and there are \(50\) divisions in its circular scale. The pitch of the screw gauge is:
1. | \(0.25\) mm | 2. | \(0.5\) mm |
3. | \(1.0\) mm | 4. | \(0.01\) mm |
1. | \(9.98\) m | 2. | \(9.980\) m |
3. | \(9.9\) m | 4. | \(9.9801\) m |
A thin wire has a length of \(21.7~\text{cm}\) and a radius of \(0.46~\text{mm}\). The volume of the wire to correct significant figures is:
1. | \( 0.15~ \text{cm}^3 \) | 2. | \( 0.1443~ \text{cm}^3 \) |
3. | \( 0.14~ \text{cm}^3 \) | 4. | \( 0.144 ~\text{cm}^3\) |
In an experiment, the height of an object measured by a vernier callipers having least count of \(0.01~\text{cm}\) is found to be \(5.72~\text{cm}\). When no object is there between jaws of this vernier callipers, the reading of the main scale is \(0.1\) cm and the reading of the vernier scale is \(0.3~\text{mm}\). The correct height of the object is:
1. \( 5.72 ~\text{cm} \)
2. \( 5.59~\text{cm} \)
3. \( 5.85~\text{cm} \)
4. \( 5.69~\text{cm} \)
In which of the following, the number of significant figures is different from that in the others?
1. | \(2.303~\text{kg}\) | 2. | \(12.23~\text{m}\) |
3. | \(0.002\times10^{5}~\text{m}\) | 4. | \(2.001\times10^{-3}~\text{kg}\) |
The sum of the numbers \(436.32,227.2,\) and \(0.301\) in the appropriate significant figures is:
1. | \( 663.821 \) | 2. | \( 664 \) |
3. | \( 663.8 \) | 4. | \(663.82\) |
The mass and volume of a body are \(4.237~\text{grams}\) and \(2.5~\text{cm}^3\), respectively. The density of the material of the body in correct significant figures will be:
1. \(1.6048~\text{grams cm}^{-3}\)
2. \(1.69~\text{grams cm}^{-3}\)
3. \(1.7~\text{grams cm}^{-3}\)
4. \(1.695~\text{grams cm}^{-3}\)
The numbers \(2.745\) and \(2.735\) on rounding off to \(3\) significant figures will give respectively,
1. | \(2.75\) and \(2.74\) | 2. | \(2.74\) and \(2.73\) |
3. | \(2.75\) and \(2.73\) | 4. | \(2.74\) and \(2.74\) |
Young's modulus of steel is \(1.9 \times 10^{11} ~\text{N/m}^2\). When expressed in CGS units of \(\text{dyne/cm}^2\), it will be equal to: \((1 \mathrm{~N}=10^5 \text { dyne, } 1~ \text{m}^2=10^4 ~\text{cm}^2)\)
1. \( 1.9 \times 10^{10} \)
2. \( 1.9 \times 10^{11} \)
3. \( 1.9 \times 10^{12} \)
4. \( 1.9 \times 10^9\)
1. | \( {\left[{pA}^{-1} {~T}^1\right]} \) | 2. | \( {\left[{p}^2 {AT}\right]} \) |
3. | \( {\left[{pA}^{-1 / 2} {~T}\right]} \) | 4. | \( {\left[{pA}^{1 / 2} {~T}^{-1}\right]}\) |