12.0 Questions

Question 1. Solve the following accordingly:

(1) Differentiate $x\sin x$ with respect to $x$

(2) Integrate $\int {{{dx} \over {4x + 5}}} $ with respect to $x$

(1) Now using the method of differentiation,

$${d \over {dx}}(x\sin x)$$ $$ =x{d \over {dx}}(\sin x) + \sin x.{d \over {dx}}(x)$$$$ = x\cos x + \sin x(1)$$$$ =x\cos x + \sin x Ans.\ $$

(2) In order to differentiate we have,

$$\int {{{dx} \over {4x + 5}}} $$ = $${1 \over 4}\int {{{dX} \over X}} {\rm{, where X = 4x + 5}}$$$$ = {1 \over 4}\ln X + {c_1}$$ $$ = {1 \over 4}\ln (4x + 5) + {c_2} $$

where ${c_1}$ and ${c_2}$ are constants of integration.

Question 2. Calculate the percentage error in the determination of time period of a pendulum.

$$T = 2\pi \sqrt {{l \over g}} $$

if given that $l$ and $g$ are measured with $ \pm 2\% $ and $ \pm 5\% $ errors.

Solution: Now we know the concept related to combination of errors so using the same,

$$\frac{{\Delta T}}{T} \times 100 = \pm \left( {\frac{1}{2} \times \frac{{\Delta l}}{l} \times 100 + \frac{1}{2} \times \frac{{\Delta g}}{g} \times 100} \right)$$$$ =\pm ({1 \over 2} \times 2 + {1 \over 2} \times 5)$$$$ =\pm3.5% Ans.\ $$

Question 3. In the diagram shown find the nature and magnitude of the zero error.

Solution: Here, zero error of vernier scale is lying to the right of zero of main scale, hence it has positive zero error.

Also,

$$N = 0,\;x = 6,\;L.C. = 0.01cm$$

Hence,

Zero error $= N + x \times V.C.$

$$=0 + 6 \times 0.01$$$$=0.06 cm$$

So, the correction applied will be,

Zero correction$ = -0.06\ cm$ Ans.

And therefore the actual length will be 0.06 cm less than the measured length.