In a screw gauge, the zero of the circular scale lies 3 divisions above the horizontal pitch line when their metallic studs are brought in contact. Using this instrument thickness of a sheet is measured. If pitch scale reading is 1 mm and the circular scale reading is 51 then the correct thickness of the sheet is_____ mm.
[Assume least count is 0.01 mm]
JEE Laboratory Experiments - XI Questions
JEE Laboratory Experiments - XI Questions
We are given a screw gauge with a zero error where the zero of the circular scale lies 3 divisions above the horizontal pitch line when the studs are in contact. The least count (LC) is $$0.01$$ mm.
When the zero of the circular scale lies 3 divisions above the reference line, the circular scale has not reached its zero mark. This means extra rotation is needed to bring it to zero, which represents a negative zero error:
$$\text{Zero error} = -3 \times \text{LC} = -3 \times 0.01 = -0.03 \text{ mm}$$
The observed reading is given by:
$$\text{Observed reading} = \text{PSR} + (\text{CSR} \times \text{LC}) = 1 + (51 \times 0.01) = 1.51 \text{ mm}$$
The zero correction is the negative of the zero error, so the corrected reading is:
$$\text{Correct thickness} = \text{Observed reading} - \text{Zero error} = 1.51 - (-0.03) = 1.54 \text{ mm}$$
The correct answer is Option B) 1.54 mm.
In a vernier callipers, 50 vernier scale divisions are equal to 48 main scale divisions. If one main scale division= 0.05 mm, then the least count of the vernier callipers is__________ mm.
We need to find the least count of the vernier callipers.
50 vernier scale divisions (VSD) = 48 main scale divisions (MSD)
1 MSD = 0.05 mm
Find the value of 1 VSD:
$$1 \text{ VSD} = \frac{48}{50} \text{ MSD} = \frac{48}{50} \times 0.05 = 0.048 \text{ mm}$$
Calculate the least count:
$$\text{Least Count} = 1 \text{ MSD} - 1 \text{ VSD} = 0.05 - 0.048 = 0.002 \text{ mm}$$
Therefore, the least count is Option 3: 0.002 mm.
Surface tension of two liquids (having same densities), $$T_{1}$$ and $$T_{2}$$, are measured using capillary rise method utilizing two tubes with inner radii of $$r_{1}$$ and $$r_{2}$$
where $$r_{1} > r_{2}$$. The measured liquid heights in these tubes are $$h_{1}$$ and $$h_{2}$$ respectively. [Ignore the weight of the liquid about the lowest point of miniscus]. The heights $$h_{1}$$ and $$h_{2}$$ and surfaces tensions $$T_{1}$$ and $$T_{2}$$ satisfy the relation :
Two liquids with the same densities and surface tensions $$T_1$$ and $$T_2$$ rise to heights $$h_1$$ and $$h_2$$, respectively, in capillary tubes of radii $$r_1 > r_2$$.
We first recall the capillary rise formula:
$$h = \frac{2T\cos\theta}{\rho g r}$$
Assuming that both liquids have the same contact angle (so that $$\cos\theta = 1$$) and the same density, we can rearrange to find
$$T = \frac{h\rho g r}{2}$$, which shows that $$T \propto hr$$.
Because the surface tensions are equal ($$T_1 = T_2$$), it follows that
$$h_1 r_1 = h_2 r_2$$, and since $$r_1 > r_2$$, we have $$h_1 < h_2$$.
This conclusion matches Option C, which states $$h_1 < h_2$$ and $$T_1 = T_2$$.
The correct answer is Option C.
When both jaws of vernier callipers touch each other, zero mark of the vernier scale is right to zero mark of main scale $$4^{th}$$ mark on vernier scale coincides with certain mark on the main scale. while measuring the length of a cylinder, observer observes 15 divisions on main scale and $$5^{th}$$ division of vernier scale coincides with a main scale division. Measured length of cylinder is______mm.
(Least count of Vernier calliper= 0.1 mm)
The reading shown by a vernier calliper has two parts:
• Main-scale reading (MSR) = value at the mark just left of the vernier zero.
• Vernier-scale reading (VSR) = (coinciding vernier division) $$\times$$ least count (LC).
Total reading $$=$$ MSR $$+$$ VSR.
Case 1: Determining the zero error (jaws closed)
When the jaws are in contact, the vernier zero is to the right of the main-scale zero; this is a positive zero error.
Here the $$4^{\text{th}}$$ vernier division coincides with a main-scale mark.
MSR = $$0 \text{ mm}$$ because the vernier zero has not crossed the first main-scale division.
VSR $$= 4 \times LC = 4 \times 0.1 \text{ mm} = 0.4 \text{ mm}$$.
Observed zero reading $$= 0 + 0.4 = 0.4 \text{ mm}$$, so
zero error $$= +0.4 \text{ mm}$$ and the zero correction (to be applied later) is $$-0.4 \text{ mm}$$.
Case 2: Measuring the cylinder
The observer notes:
• MSR = $$15 \text{ mm}$$ (15 main-scale divisions lie to the left of the vernier zero).
• The $$5^{\text{th}}$$ vernier division coincides with a main-scale mark.
VSR $$= 5 \times 0.1 \text{ mm} = 0.5 \text{ mm}$$.
Raw reading (uncorrected) $$= 15 + 0.5 = 15.5 \text{ mm}$$.
Applying the zero correction
True length $$= 15.5 \text{ mm} - 0.4 \text{ mm} = 15.1 \text{ mm}$$.
Therefore, the measured length of the cylinder is $$\mathbf{15.1 \text{ mm}}$$ (Option B).

