CBSE Class 11 Physics — NCERT Solutions

(a) Volume of a cube of side 1 cm in m³:

Given: side $= 1\,\text{cm} = 1 \times 10^{-2}\,\text{m}$

$$V = (1\times10^{-2})^3 = 10^{-6}\,\text{m}^3$$

Answer: $10^{-6}\,\text{m}^3$

(b) Surface area of a solid cylinder (radius = 2.0 cm, height = 10.0 cm) in mm²:

Formula: $A = 2\pi r(r + h)$

Convert to mm: $r = 20\,\text{mm}$, $h = 100\,\text{mm}$

$$A = 2\pi \times 20 \times (20 + 100) = 2\pi \times 20 \times 120$$
$$A = 2 \times 3.14159 \times 2400 = 15079.6\,\text{mm}^2$$
$$A \approx 1.5 \times 10^4\,\text{mm}^2$$

Answer: $\approx 1.5 \times 10^4\,\text{mm}^2$

(c) Distance covered in 1 s at 18 km h⁻¹:

$$18\,\text{km h}^{-1} = 18 \times \frac{1000\,\text{m}}{3600\,\text{s}} = 5\,\text{m s}^{-1}$$

Distance in 1 s $= 5 \times 1 = 5\,\text{m}$

Answer: $5\,\text{m}$

(d) Density of lead:

Relative density (specific gravity) $= \dfrac{\text{density of substance}}{\text{density of water}}$

Density of water $= 1\,\text{g cm}^{-3} = 1000\,\text{kg m}^{-3}$

$$\rho_{\text{lead}} = 11.3 \times 1\,\text{g cm}^{-3} = 11.3\,\text{g cm}^{-3}$$
$$\rho_{\text{lead}} = 11.3 \times 1000\,\text{kg m}^{-3} = 1.13 \times 10^4\,\text{kg m}^{-3}$$

Answer: $11.3\,\text{g cm}^{-3}$ or $1.13 \times 10^4\,\text{kg m}^{-3}$

(a) $1\,\text{kg m}^2\,\text{s}^{-2}$ in g cm² s⁻²:

$1\,\text{kg} = 10^3\,\text{g}$, $1\,\text{m} = 10^2\,\text{cm}$

$$1\,\text{kg m}^2\,\text{s}^{-2} = 10^3\,\text{g} \times (10^2\,\text{cm})^2 \times \text{s}^{-2}$$
$$= 10^3 \times 10^4\,\text{g cm}^2\,\text{s}^{-2} = 10^7\,\text{g cm}^2\,\text{s}^{-2}$$

Answer: $10^7\,\text{g cm}^2\,\text{s}^{-2}$

(b) 1 m in light years (ly):

Speed of light $c = 3 \times 10^8\,\text{m s}^{-1}$

1 year $= 365.25 \times 24 \times 3600\,\text{s} \approx 3.156 \times 10^7\,\text{s}$

$1\,\text{ly} = 3 \times 10^8 \times 3.156 \times 10^7 = 9.467 \times 10^{15}\,\text{m}$

$$1\,\text{m} = \frac{1}{9.467 \times 10^{15}}\,\text{ly} \approx 1.057 \times 10^{-16}\,\text{ly}$$

Answer: $\approx 1.057 \times 10^{-16}\,\text{ly}$

(c) $3.0\,\text{m s}^{-2}$ in km h⁻²:

$1\,\text{m} = 10^{-3}\,\text{km}$, $1\,\text{s} = \dfrac{1}{3600}\,\text{h}$, so $1\,\text{s}^{-1} = 3600\,\text{h}^{-1}$

$$3.0\,\text{m s}^{-2} = 3.0 \times 10^{-3}\,\text{km} \times (3600)^2\,\text{h}^{-2}$$
$$= 3.0 \times 10^{-3} \times 1.296 \times 10^7\,\text{km h}^{-2}$$
$$= 3.888 \times 10^4\,\text{km h}^{-2} \approx 3.9 \times 10^4\,\text{km h}^{-2}$$

Answer: $3.9 \times 10^4\,\text{km h}^{-2}$

(d) $G = 6.67 \times 10^{-11}\,\text{N m}^2\,\text{kg}^{-2}$ in cm³ s⁻² g⁻¹:

Note: $1\,\text{N} = 1\,\text{kg m s}^{-2}$, so $\text{N m}^2\,\text{kg}^{-2} = \text{kg m s}^{-2} \cdot \text{m}^2 \cdot \text{kg}^{-2} = \text{m}^3\,\text{s}^{-2}\,\text{kg}^{-1}$

Convert: $1\,\text{m}^3 = 10^6\,\text{cm}^3$, $1\,\text{kg}^{-1} = (10^3\,\text{g})^{-1} = 10^{-3}\,\text{g}^{-1}$

$$G = 6.67 \times 10^{-11} \times 10^6\,\text{cm}^3 \times \text{s}^{-2} \times 10^{-3}\,\text{g}^{-1}$$
$$= 6.67 \times 10^{-11} \times 10^3\,\text{cm}^3\,\text{s}^{-2}\,\text{g}^{-1}$$
$$= 6.67 \times 10^{-8}\,\text{cm}^3\,\text{s}^{-2}\,\text{g}^{-1}$$

Answer: $6.67 \times 10^{-8}\,\text{cm}^3\,\text{s}^{-2}\,\text{g}^{-1}$

For the same load, the extension is the same:

$$\Delta L=\frac{FL}{AY}$$

So for steel and copper,

$$\frac{F L_s}{A_s Y_s}=\frac{F L_c}{A_c Y_c}$$

Cancel $F$:

$$\frac{Y_s}{Y_c}=\frac{L_s A_c}{L_c A_s}$$

Substitute the values:

$$\frac{Y_s}{Y_c}=\frac{4.7\times 4.0\times 10^{-5}}{3.5\times 3.0\times 10^{-5}}$$

$$=\frac{4.7\times 4.0}{3.5\times 3.0}=\frac{18.8}{10.5}\approx 1.79$$

So the ratio of Young's moduli is about $1.8$.

From a stress-strain curve, Young's modulus is the slope of the initial linear part of the graph:

$$Y=\frac{\text{stress}}{\text{strain}}$$

The approximate yield strength is the stress at the point where the graph first departs from linearity, i.e. the yield point.

So, read the slope of the straight-line part for $Y$ and the stress at the yield point for the yield strength.

A body can be treated as a point object when its size is much smaller than the distance it travels, i.e., the internal motion or rotation of the body is irrelevant to the problem.

(a) Railway carriage moving between two stations — YES, it can be treated as a point object.
The size of the carriage (~20 m) is negligible compared to the distance between two stations (several kilometres). The motion is smooth (no jerks), so internal details are unimportant.

(b) Monkey sitting on top of a man cycling on a circular track — YES, it can be treated as a point object.
The size of the monkey is negligible compared to the size of the circular track. The monkey is sitting still relative to the man, so no internal motion matters.

(c) Spinning cricket ball that turns sharply on hitting the ground — NO, it cannot be treated as a point object.
The spin (rotation) of the ball is crucial to understanding its sharp turn. The size and rotational motion of the ball are important here.

(d) Tumbling beaker that has slipped off the edge of a table — NO, it cannot be treated as a point object.
The beaker is tumbling (rotating), so its orientation and rotational motion are significant. It cannot be reduced to a point.

From the x–t graph (Fig. 2.9), we read the following information:

• The x-axis represents position (distance from school O) and the t-axis represents time.
• A's home P is at a smaller distance from school than B's home Q.
• B's graph starts from the origin (school) at an earlier time than A's graph.
• The slope of the x–t graph gives speed. B's line has a steeper slope than A's line.
• Both graphs end (reach home) at the same time.
• The two lines intersect once, meaning one child overtakes the other once.

(a) A lives closer to the school than B.
(Home P of A is at a smaller x-value than home Q of B on the graph.)

(b) B starts from the school earlier than A.
(B's graph begins at an earlier time on the t-axis.)

(c) B walks faster than A.
(The slope of B's x–t graph is steeper than A's, indicating greater speed.)

(d) A and B reach home at the same time.
(Both graphs terminate at the same value of t.)

(e) B overtakes A on the road once.
(The two lines cross once, meaning B, who started earlier but walks faster, overtakes A once.)

(a) Blood pressure greater at feet than at brain:

The pressure in a fluid at rest increases with depth according to $P = P_0 + \rho g h$. The feet are at a much greater vertical distance (height $h$) below the heart compared to the brain. Therefore, the blood pressure at the feet is greater than at the brain by an amount $\rho g h$, where $h$ is the height difference between the brain and the feet.

(b) Atmospheric pressure halves at ~6 km:

The atmosphere is not a liquid of uniform density. The density of air decreases rapidly with altitude. Most of the mass of the atmosphere is concentrated in the lower layers (near the Earth's surface). Since pressure at any point equals the weight of the air column above it per unit area, and since most of the atmospheric mass lies below 6 km, the pressure at 6 km is nearly half the sea-level pressure. The upper layers (above 6 km) are very thin and contribute little to the total pressure.

(c) Hydrostatic pressure is a scalar:

Although pressure is defined as force per unit area ($P = F/A$), the force here is the normal force exerted by the fluid on a surface. When we say pressure at a point in a fluid, it acts equally in all directions (Pascal's law). There is no unique direction associated with pressure at a point — it has the same magnitude regardless of the orientation of the surface element chosen. Hence, hydrostatic pressure is a scalar quantity.

Given: A list of physical quantities.

Concept: Scalar quantities have only magnitude; vector quantities have both magnitude and direction.

Explanation:

• Scalars: Volume, mass, speed, density, number of moles, and angular frequency are completely described by their magnitude alone.
• Vectors: Acceleration, velocity, displacement, and angular velocity require both magnitude and direction for complete description.

Given:

• Triple point of Neon: $T_{Ne} = 24.57\,\text{K}$
• Triple point of Carbon dioxide: $T_{CO_2} = 216.55\,\text{K}$

Formula used:
$$t_C = T - 273.15$$
$$t_F = \frac{9}{5}\,t_C + 32$$

For Neon:

Celsius scale:
$$t_C = 24.57 - 273.15 = -248.58^\circ\text{C}$$

Fahrenheit scale:
$$t_F = \frac{9}{5}\times(-248.58) + 32 = -447.44 + 32 = -415.44^\circ\text{F}$$

For Carbon dioxide:

Celsius scale:
$$t_C = 216.55 - 273.15 = -56.60^\circ\text{C}$$

Fahrenheit scale:
$$t_F = \frac{9}{5}\times(-56.60) + 32 = -101.88 + 32 = -69.88^\circ\text{F}$$

Results:

• Neon: $-248.58^\circ\text{C}$, $-415.44^\circ\text{F}$
• Carbon dioxide: $-56.60^\circ\text{C}$, $-69.88^\circ\text{F}$

Concept: Newton's Second Law — if acceleration $a = 0$, net force $F = ma = 0$.

(a) Rain drop falling with constant speed:
Since speed is constant, acceleration $a = 0$.
$$F_{net} = ma = 0 \text{ N}$$
The net force is zero.

(b) Cork of mass 10 g floating on water:
The cork is in equilibrium (stationary), so $a = 0$.
$$F_{net} = 0 \text{ N}$$
The net force is zero.

(c) Kite held stationary in the sky:
The kite is stationary, so $a = 0$.
$$F_{net} = 0 \text{ N}$$
The net force is zero.

(d) Car moving with constant velocity of 30 km/h:
Constant velocity means $a = 0$.
$$F_{net} = 0 \text{ N}$$
The net force is zero.

(e) High-speed electron in space, free of all fields:
No gravitational, electric, or magnetic force acts on it, so:
$$F_{net} = 0 \text{ N}$$
The net force is zero. The electron moves with constant velocity (Newton's First Law).

Given:

• Flow rate of water = 3.0 litres/min = 3.0 kg/min (since density of water = 1 kg/litre)
• Initial temperature, $T_1 = 27°C$
• Final temperature, $T_2 = 77°C$
• Rise in temperature, $\Delta T = 77 - 27 = 50°C$
• Heat of combustion of fuel, $H = 4.0 \times 10^4$ J/g
• Specific heat of water, $s = 4.2 \times 10^3$ J kg⁻¹ K⁻¹

Concept: Heat required = $ms\Delta T$

Step 1: Calculate heat required per minute
$$\Delta Q = ms\Delta T = 3.0 \times 4.2 \times 10^3 \times 50$$
$$\Delta Q = 6.3 \times 10^5 \text{ J/min}$$

Step 2: Calculate rate of fuel consumption

Let the rate of fuel consumption be $r$ g/min.

Heat supplied by fuel per minute = $r \times H = r \times 4.0 \times 10^4$ J/min

Setting heat supplied equal to heat required:
$$r \times 4.0 \times 10^4 = 6.3 \times 10^5$$
$$r = \frac{6.3 \times 10^5}{4.0 \times 10^4} = 15.75 \text{ g/min}$$

Answer: The rate of consumption of fuel is approximately 15.75 g/min.

The work done by a force is given by $W = F \cdot d \cos\theta$, where $\theta$ is the angle between the force and displacement.

(a) Positive.
The man applies force upward (via rope) and the bucket moves upward. The force and displacement are in the same direction ($\theta = 0°$), so work done is positive.

(b) Negative.
Gravitational force acts downward but the bucket moves upward. The force and displacement are in opposite directions ($\theta = 180°$), so work done by gravity is negative.

(c) Negative.
Friction on a body sliding down an inclined plane acts up the plane (opposing motion), while displacement is down the plane. Since $\theta = 180°$, work done by friction is negative.

(d) Positive.
The applied force is in the direction of motion (horizontal). Even though friction also acts, the applied force itself does positive work ($\theta = 0°$).

(e) Negative.
The resistive force of air opposes the motion of the pendulum at every point. Since force and displacement are always in opposite directions ($\theta = 180°$), work done by air resistance is negative.

For a gas, the fraction of molecular volume to total volume is proportional to the density compared with liquid water. From the chapter’s Example 12.1, for water vapour at $100^\circ\mathrm{C}$ and 1 atm this fraction was estimated as about $6\times10^{-4}$. Using the same method for oxygen at STP, the order is the same: the molecules occupy only a very small fraction of the total volume.

So the estimated fraction is $6\times10^{-4}$.

For a homogeneous body, symmetry shows that the mass is distributed equally about the geometric centre. Hence the centre of mass lies at the geometric centre for a sphere, cylinder, ring and cube.

The centre of mass need not always be inside the body. In a ring, it is at the centre, which lies outside the material of the ring.

Examples (a), (b), and (c) represent periodic motion.

• (a) A swimmer returning to the same bank and back repeats the motion.
• (b) A freely suspended bar magnet displaced and released oscillates periodically.
• (c) A hydrogen molecule rotating about its centre of mass repeats its position after every revolution.
• (d) An arrow released from a bow does not repeat its motion, so it is not periodic.

No. Gravitational shielding is not possible. A hollow sphere does not block gravity the way a hollow conductor can block electric forces. The mass of the shell still attracts bodies inside and outside it, and for a body inside a uniform spherical shell the net gravitational force is zero only because the attractions cancel, not because the gravity is shielded. So there is no practical way to shield a body from nearby gravitational influence by putting it inside a hollow sphere or by other means.

Given:

• Mass of string, $m = 2.50$ kg
• Tension, $T = 200$ N
• Length, $L = 20.0$ m

Formula used:
Speed of transverse wave on a string: $$v = \sqrt{\frac{T}{\mu}}$$
where $\mu$ = linear mass density = $\dfrac{m}{L}$

Step 1: Find linear mass density
$$\mu = \frac{m}{L} = \frac{2.50}{20.0} = 0.125 \text{ kg m}^{-1}$$

Step 2: Find speed of wave
$$v = \sqrt{\frac{T}{\mu}} = \sqrt{\frac{200}{0.125}} = \sqrt{1600} = 40 \text{ m s}^{-1}$$

Step 3: Find time to travel length $L$
$$t = \frac{L}{v} = \frac{20.0}{40} = 0.50 \text{ s}$$

Answer: The disturbance takes 0.50 s to reach the other end.