Numerical Problems Based on Class 11 Physics Units and Measurement

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Home CBSE Class 11 Physics Extra Questions for Class 11 Physics Numerical Problems Based on Class 11 Physics Units and Measurement

Numerical Problems on Units and Measurement

Class 11 Physics · Units and Measurement

Units and Measurement Practice Set 1

Apply dimensional analysis and parallax methods

Ch 2 · Units & Measurement

Convert a power of one mega watt on a system whose fundamental units are 10 kg, 1 dm and 1 minute.

Answer & Solution

Answer: $2.16 \times 10^{12}$ in the new units

Table of Contents

Explanation: The dimensional formula for power is $[M^1 L^2 T^{-3}]$.
In the SI system: $n_1 = 1\text{ MW} = 10^6\text{ W}$. $M_1 = 1\text{ kg}$, $L_1 = 1\text{ m}$, $T_1 = 1\text{ s}$.
In the new system: $M_2 = 10\text{ kg}$, $L_2 = 1\text{ dm} = 0.1\text{ m}$, $T_2 = 1\text{ min} = 60\text{ s}$.

Using the conversion formula $n_2 = n_1 \left[\frac{M_1}{M_2}\right]^a \left[\frac{L_1}{L_2}\right]^b \left[\frac{T_1}{T_2}\right]^c$:

$$n_2 = 10^6 \left[\frac{1\text{ kg}}{10\text{ kg}}\right]^1 \left[\frac{1\text{ m}}{0.1\text{ m}}\right]^2 \left[\frac{1\text{ s}}{60\text{ s}}\right]^{-3}$$

$$n_2 = 10^6 \times \left(\frac{1}{10}\right) \times (10)^2 \times (60)^3$$

$$n_2 = 10^6 \times 0.1 \times 100 \times 216000 = 10^7 \times 216000 = 2.16 \times 10^{12}$$

The moon is observed from two diametrically opposite points A and B on earth. The angle $\theta$ subtended at the moon by the two directions of observation is $1^\circ54’$. Calculate the distance of moon from earth. [Given the diameter of earth = $1.276 \times 10^7\text{ m}$]

Answer & Solution

Answer: $3.84 \times 10^8\text{ m}$

Explanation: This is based on the parallax method where $D = \frac{b}{\theta}$.
Basis $b$ (diameter of earth) $= 1.276 \times 10^7\text{ m}$.
Parallax angle $\theta = 1^\circ54′ = 60′ + 54′ = 114’$.

First, convert $\theta$ into radians. Since $1′ = 2.91 \times 10^{-4}\text{ rad}$:

$$\theta = 114 \times 2.91 \times 10^{-4}\text{ rad} = 3.32 \times 10^{-2}\text{ rad}$$

$$D = \frac{1.276 \times 10^7\text{ m}}{3.32 \times 10^{-2}\text{ rad}} \approx 3.84 \times 10^8\text{ m}$$

The velocity of sound in air is 332 m/s. If the units of length is km and unit of time is hour, using dimensions calculate the value of velocity?

Answer & Solution

Answer: $1195.2\text{ km/h}$

Explanation: The dimensional formula for velocity is $[L^1 T^{-1}]$.
Given $n_1 = 332$, $L_1 = 1\text{ m}$, $T_1 = 1\text{ s}$.
Target $L_2 = 1\text{ km} = 1000\text{ m}$, $T_2 = 1\text{ hr} = 3600\text{ s}$.

Using $n_2 = n_1 \left[\frac{L_1}{L_2}\right]^1 \left[\frac{T_1}{T_2}\right]^{-1}$:

$$n_2 = 332 \left[\frac{1\text{ m}}{1000\text{ m}}\right]^1 \left[\frac{1\text{ s}}{3600\text{ s}}\right]^{-1}$$

$$n_2 = 332 \times \frac{1}{1000} \times 3600 = 332 \times 3.6 = 1195.2$$

The angular diameter of the sun from earth is 16 minute. If the distance of earth from the sun is $1.49 \times 10^{11}\text{ m}$, then what will be the diameter of sun?

Answer & Solution

Answer: $6.93 \times 10^8\text{ m}$

Explanation: Using the relation $d = D\theta$ where $d$ is diameter, $D$ is distance, and $\theta$ is angular diameter in radians.
Distance $D = 1.49 \times 10^{11}\text{ m}$.
Angle $\theta = 16′ = 16 \times 2.91 \times 10^{-4}\text{ rad} = 4.656 \times 10^{-3}\text{ rad}$.

$$d = (1.49 \times 10^{11}) \times (4.656 \times 10^{-3}) = 6.93 \times 10^8\text{ m}$$

A physical quantity Q is given by $Q = \frac{A^2 B^3}{C \sqrt{D}}$. The percentage error in A, B, C, D are 1%, 2%, 4%, 2% respectively. Find the percentage error in Q.

Answer & Solution

Answer: $13\%$

Explanation: (Assuming the standard relation $Q = \frac{A^2 B^3}{C \sqrt{D}}$ inferred from the context)

The maximum percentage error is the sum of the individual percentage errors multiplied by their respective powers:

$$\frac{\Delta Q}{Q} \times 100 = 2\left(\frac{\Delta A}{A} \times 100\right) + 3\left(\frac{\Delta B}{B} \times 100\right) + 1\left(\frac{\Delta C}{C} \times 100\right) + \frac{1}{2}\left(\frac{\Delta D}{D} \times 100\right)$$

Substitute the given percentage errors:

$$\% \text{ Error} = 2(1\%) + 3(2\%) + 1(4\%) + 0.5(2\%)$$

$$\% \text{ Error} = 2\% + 6\% + 4\% + 1\% = 13\%$$

Units and Measurement Practice Set 2

Calculate error limits and propagate uncertainties

Ch 2 · Units & Measurement

A body has an acceleration of $10\text{ km/h}^2$. Find its value in c.g.s system.

Answer & Solution

Answer: $0.0772\text{ cm/s}^2$

Explanation: In the c.g.s system, acceleration is measured in $\text{cm/s}^2$.
$1\text{ km} = 100000\text{ cm} = 10^5\text{ cm}$.
$1\text{ hour} = 3600\text{ s}$.

$$a = 10 \frac{\text{km}}{\text{h}^2} = 10 \times \frac{10^5\text{ cm}}{(3600\text{ s})^2}$$

$$a = \frac{10^6}{12960000}\text{ cm/s}^2 = \frac{100}{1296}\text{ cm/s}^2 \approx 0.07716\text{ cm/s}^2$$

The resistance R is the ratio of potential difference V and current I. What is the % error in R if V = ($100 \pm 5$) volt and I = ($10 \pm 0.2$) A?

Answer & Solution

Answer: $7\%$

Explanation: Given $R = \frac{V}{I}$. The maximum percentage error in $R$ is the sum of the percentage errors in $V$ and $I$.

$$\frac{\Delta R}{R} \times 100 = \left(\frac{\Delta V}{V} \times 100\right) + \left(\frac{\Delta I}{I} \times 100\right)$$

$$\% \text{ Error} = \left(\frac{5}{100} \times 100\right) + \left(\frac{0.2}{10} \times 100\right)$$

$$\% \text{ Error} = 5\% + 2\% = 7\%$$

Two resistances $r_1 = (5.0 \pm 0.2)\text{ }\Omega$ and $r_2 = (10.0 \pm 0.1)\text{ }\Omega$ are connected in parallel. Find the value of equivalent resistance with limits of percentage error.

Answer & Solution

Answer: $(3.33 \pm 0.1)\text{ }\Omega$

Explanation: First, calculate the equivalent parallel resistance $R_p$:

$$R_p = \frac{r_1 r_2}{r_1 + r_2} = \frac{5.0 \times 10.0}{5.0 + 10.0} = \frac{50}{15} \approx 3.33\text{ }\Omega$$

The error in parallel resistance is given by differentiating $\frac{1}{R_p} = \frac{1}{r_1} + \frac{1}{r_2}$:

$$\frac{\Delta R_p}{R_p^2} = \frac{\Delta r_1}{r_1^2} + \frac{\Delta r_2}{r_2^2}$$

$$\Delta R_p = R_p^2 \left( \frac{0.2}{(5.0)^2} + \frac{0.1}{(10.0)^2} \right) = (3.33)^2 \left( \frac{0.2}{25} + \frac{0.1}{100} \right)$$

$$\Delta R_p = 11.11 \times (0.008 + 0.001) = 11.11 \times 0.009 \approx 0.1\text{ }\Omega$$

So, $R_p = 3.3 \pm 0.1\text{ }\Omega$. To find the percentage error limit: $\frac{0.1}{3.33} \times 100 \approx 3\%$.

Calculate the focal length of a spherical mirror from the following data: Object distance $u = (50.1 \pm 0.5)\text{ cm}$, image distance $v = (20 \pm 0.2)\text{ cm}$.

Answer & Solution

Answer: $(14.3 \pm 0.14)\text{ cm}$

Explanation: Using the mirror formula $\frac{1}{f} = \frac{1}{v} + \frac{1}{u}$:

$$f = \frac{uv}{u+v} = \frac{50.1 \times 20}{50.1 + 20} = \frac{1002}{70.1} \approx 14.29\text{ cm}$$

To find the error, use the differentiated form $\frac{\Delta f}{f^2} = \frac{\Delta u}{u^2} + \frac{\Delta v}{v^2}$:

$$\Delta f = f^2 \left( \frac{0.5}{(50.1)^2} + \frac{0.2}{(20)^2} \right) = (14.29)^2 \left( \frac{0.5}{2510} + \frac{0.2}{400} \right)$$

$$\Delta f = 204.2 \times (0.0002 + 0.0005) = 204.2 \times 0.0007 \approx 0.14\text{ cm}$$

A capacitor of capacitance $C = (2.0 \pm 0.1)\text{ }\mu\text{F}$ is charged to a voltage $V = (20 \pm 0.5)\text{ V}$. Calculate the charge Q with error limits.

Answer & Solution

Answer: $(40 \pm 3)\text{ }\mu\text{C}$

Explanation: The charge is given by the formula $Q = CV$.

$$Q = 2.0\text{ }\mu\text{F} \times 20\text{ V} = 40\text{ }\mu\text{C}$$

For a product, the fractional errors are added:

$$\frac{\Delta Q}{Q} = \frac{\Delta C}{C} + \frac{\Delta V}{V} = \frac{0.1}{2.0} + \frac{0.5}{20}$$

$$\frac{\Delta Q}{Q} = 0.05 + 0.025 = 0.075$$

Now, solve for absolute error $\Delta Q$:

$$\Delta Q = 40 \times 0.075 = 3.0\text{ }\mu\text{C}$$

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