A student measured the length of a rod and wrote its as \(3.50~\text{cm}\). Which instrument did he use to measure it?

A student measures the time period of \(100\) oscillations of a simple pendulum four times. The data set is \(90~\text{s}, ~91~\text{s},~95~\text{s}~\text{and}~92~\text{s}.\) If the minimum division in the measuring clock is \(1~\text{s}\), then the reported mean time should be:
1. \( 92 \pm 2 ~\text{s} \)
2. \( 92 \pm 5.0 ~\text{s} \)
3. \( 92 \pm 1.8 ~\text{s} \)
4. \( 92 \pm 3~\text{s} \)

A screw gauge with a pitch of \(0.5~\text{mm}\) and a circular scale with \(50\) divisions is used to measure the thickness of a thin sheet of Aluminium. Before starting the measurement, it is found that when the two jaws of the screw gauge are brought in contact, the \(45^{\text{th}}\) division coincides with the main scale line and that the zero of the main scale is barely visible. What is the thickness of the sheet if the main scale reading is \(0.5~\text{mm}\) and the \(25^{\text{th}}\) division coincides with the main scale line?
1. \(0.75~\text{mm}\)
2. \(0.80~\text{mm}\)
3. \(0.70~\text{mm}\)
4. \(0.50~\text{mm}\)

The least count of the main scale of a vernier callipers is \(1~\mathrm{mm}\). Its vernier scale is divided into \(10\) divisions and coincide with \(9\) divisions of the main scale. When jaws are touching each other, the \(7^{th}\) division of vernier scale coincides with a division of main scale and the zero of vernier scale is lying right side of the zero of main scale. When this vernier is used to measure length of a cylinder the zero of the vernier scale between \(3.1~\mathrm{cm}\) and \(3.2~\mathrm{cm}\) and \(4^{th}\) VSD coincides with a main scale division. The length of the cylinder is : (VSD is vernier scale division)
1. \(3.21~\text{cm}\)
2. \(2.99 ~\text{cm}\) 
3. \(3.2~\text{cm}\)
4. \(3.07~\text{cm}\)

Using a screw gauge with pitch \(0.1 ~\text{cm}\) and \(50\) divisions on its circular scale, the thickness of an object is measured. It should be accurately recorded as:
1. \(2.124~\text{cm}\)
2. \(2.121~\text{cm}\)
3. \(2.125~\text{cm}\)
4. \(2.123~\text{cm}\)

A screw gauge has \(50\) divisions on its circular scale. The circular scale is \(4\) units ahead of the pitch scale marking, prior to use. Upon one complete rotation of the circular scale, a displacement of \(0.5~\text{mm}\) is noticed on the pitch scale. The nature of the zero error involved and the last count of the screw gauge are respectively:
1. Positive, \(10~\mu\text{m}\)
2. Negative, \(2~\mu\text{m}\)
3. Positive, \(0.1~\mu\text{m}\)
4. Positive, \(0.1~\mu\text{m}\)

A student measuring the diameter of a pencil of circular cross- section with the help of a vernier scale records the following four readings  \(5.50 ~\text{mm}\), \(5.55~\text{mm}\), \(5.45~\text{mm}\), \(5.65~\text{mm}\). The average of these four readings is \(5.5375~\text{mm}\) and the standard deviation of the data is \(0.07395~\text{mm}\). The average diameter of the pencil should therefore be recorded as: 

The pitch of the screw gauge is \(1\) mm and there are \(100\) divisions on the circular scale. When nothing is put in between the jaws, the zero of the circular scale lies \(8\) divisions below the reference line. When a wire is placed between the jaws, the first linear scale division is clearly visible while the \(72\)^nd division on a circular scale coincides with the reference line. The radius of the wire is:
1. \(1.64\) mm
2. \(0.82\) mm
3. \(1.80\) mm
4. \(0.90\) mm