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Group A
1. A meter rule is used to measure the length of a piece of string in a certain experiment. It is found to be 20 cm long to the nearest millimetre. How should this result be recorded in a table of results?
Ans: b is the correct answer.
solution:
1mm = 0.1 cm = 0.001 m
measurement = 20cm = 0.200m
(since the meter rule can read to the nearest millimeter, 3 significant figures are necessary.)
2. Forces are applied to a rigid body. The forces all act in the same plane. In which diagram is the body in equilibrium?
Ans: B is the correct answer. In option b, the downward force balances the upward force.
3. An athlete makes a long jump and follows a projectile motion. Air resistance is negligible. Which one of the following statements is true about the athlete?
Ans: b. The athlete has a constant horizontal velocity and constant downward acceleration.
Explanation: In projectile motion, the horizontal component of the body remains constant throughout the motion and acceleration due to gravity always acts towards the center of the earth.
4. At Kulekhani-I Hydro-power station, water flows from Indra Sarowar into the turbines that are a vertical distance of 550 m below the lake, as shown in the diagram. Generally, 780 000 kg of water flows into the turbines every minute. The turbines have an efficiency of 85%. What is the output power of the turbines?
Ans: Option b (60 MW) is the correct answer.
5. Graphs of stress-strain for four different materials are shown below. Which graph represents the stiffest material?
Ans: a is the correct ans.
6. A boy walks towards a stationary plane mirror at a speed of 1.2 ms-1. What is the relative speed of approach of the boy and his image?
(a) zero (b) 1.2 ms-1 (c) 2.4 ms-1 (d) 1.44 ms-1
Ans: c is the correct ans.
\begin{align*} v_{r} &= u-(-v)\\ &= 1.2 - (-1.2)\\ \therefore\ v_{r} &= 2.4\ m/s. \end{align*}7. The critical angle between an equilateral prism and air is 45o. What happens to the incident ray perpendicular to the refracting surface?
Ans: a. It is reflected totally from the second surface and emerges perpendicular from the third surface.
8. In the formation of a rainbow, the light from the sun on water droplets undergoes which of the following phenomenon/phenomena?
Ans: (c) dispersion and total internal reflection
9. In what unit is the power of lens measured?
Ans: (c) dioptre.
10. A piece of wire of resistance R is bent through 180o at mid-point and the two halves are twisted together. What is the resistance of the wire thus formed?
Ans: (a) R/4
11. What are the elementary particles with half-spin called?
Ans: (c) fermions.
Group B
1. (a) State the law of conservation of momentum.
Ans: It states that “if there is no external force applied on the system, the total linear momentum of the body remains conserved. i.e total linear momentum before collision = total linear momentum after the collision.”
Mathematical expression:
m_{1}u_{1} + m_{2}u_{2} = m_{1}v_{1} + m_{2}v_{2}(b) A jumbo jet of mass 4×105 kg travelling at a speed of 5000 m/s lands on the airport. It takes 2 minutes to come to rest. Calculate the average force applied by the ground on the aeroplane.
solution:
Given:
mass (m) = 4×105
speed (v) = 5000 m/s
time, (t) = 2 min = 120 sec
Average force on the aeroplane (Favg) =?
\begin {align*} F_{avg} &= \frac{change\ in\ momentum}{time}\\ &= \frac{mv-mu}{t}\\ &= -\frac{4 \times 10^{5} \times 5000}{120}\ \left ( \because v = 0 \right )\\ \therefore F_{avg} &= -1.67 \times 10^{7}N \end{align*}OR
a. State Hook’s law.
Ans: It states that “The restoring force acting on a body is directly proportional to the extension produced within the elastic limit. i.e. F α e.
b. The walls of the tyres on a car are made of a rubber compound. The variation with the stress of the strain of a specimen of this rubber compound is shown in Fig. 1.2.
Ans: The figure given above shows the curve of the ‘elastic hysteresis’ phenomenon. The area between the curve gives the total energy lost as heat during the loading-unloading cycle. So, tyre becomes warm.
2. (a) what is meant by the specific latent heat of vaporization of water = 2.26MJkg−1?
Ans: It means that 2.26 MJ of heat energy is required to vaporize 1 kg mass of water from liquid to vapor state at its boiling point.
(b) A 1.0kW kettle contains 500g of boiling water. Calculate the time needed to evaporate all the water in the kettle. (Specific latent heat of vaporization of water = 2.26MJkg−1).
Solution:
Power of a heater (P)= 1 kW = 1000 W
Mass of boiling water (m)= 500 g = 0.5 kg
Specific latent heat of vaporization of water (L)= 2.26 MJ/kg = 2.26×106 J/kg
Time needed to evaporate all the water, t =?
Now,
Heat required to convert water to vapor is:
\begin{align*} Q &= mL\\ &= 0.5 \times 2.26 \times 10^{6}\\ &= 1.13 \times 10^{6}\ J\\ Again,\\ P&=\frac{Q}{t}\\ or,\ 1000 &= \frac{1.13 \times 10^{6}}{t}\\ or,\ t &= \frac{1.13 \times 10^{6}}{1000}\\ \therefore\ t &= 1130\ sec. \end{align*}(c) Explain why the actual time needed is a little longer than the time calculated in 2(b).
Ans: The actual time needed is longer than calculated because the heat that we provide is not directly passed to the water. It passes through the surface of the vessel in which water is being heated. So, it takes a little longer time than the calculated time.
3. (a) State any three properties of an ideal gas as assumed by the kinetic theory of gas.
Ans: Any three properties of an ideal gas as assumed by the kinetic theory of gas are:
i. The gas molecules behave as rigid, elastic, and smooth spheres.
ii. The size of gas molecules is small in comparison to the average distance between particles and the dimensions of the container.
iii. The molecules collide with one another and also with the walls of the container. These collisions are perfectly elastic.
(b) A student needed to use the ideal gas for a certain experiment. But, the ideal gas does not exist. Suggest what two different things this student could do to solve his problem.
Ans: An ideal gas is a theoretical gas whose molecules occupy negligible space and have interactions only by elastic nature. So, the student can lower the pressure and increase the temperature of the real gas which can quantitatively behave like an ideal gas.
4. (a) Define temperature gradient in an object.
Ans: Temperature gradient is defined as the rate of change of temperature with the distance between hot and cold surfaces. If dθ temperature difference occurs for distance dx, the temperature gradient is given by:
Temperature\ gradient = \frac{d\theta}{dx}(b) An electric kitchen range has a total wall area of 1.40 m2 and is insulted with a layer of fiber glass that has a temperature of 175°C and its outside surface is 35 °C. The fiber glass has a thermal conductivity of 0.040 Wm-1K-1. Calculate the rate of flow of heat through the insulation, assuming the fibre as a flat slab of area of 1.40 m2.
Solution:
Given,
Area of fiber glass (A) = 1.40 m2
Thickness of fiber glass (x) = ?? value not given…
Internal temperature (θ1) = 1750 C
External temperature (θ2) = 350 C
Thermal conductivity of fiber glass (k) = 0.040 Wm-1K-1
Rate of heat flow (Q/t) =?
Now,
To calculate the rate of heat flow, we can use the given formula:
\frac{Q}{t}= \frac{kA(\theta_{2}-\theta_{1})}{x}(c) How might the rate of conduction be affected if the fiber absorbs moisture? Justify your answer.
Ans: The thermal conductivity of the fiber will increase with an increase in moisture.
5. Figure 5.1 shows a ray of light is entering and emerging through a part of a convex lens.
(i) Define ‘convex lens’, and state one daily application of it.
Ans: Lens which is thick at the center and thin at the edges and converge the rays of light passing through it is called a convex lens. As it converges the ray of light, it is also known as a converging lens. one of the most common uses of the convex lens is in magnifying glasses, eg a camera.
(ii) Explain why this lens is also called a converging lens?
Ans: when parallel light rays pass through the lens, the refracted rays converge at one point. So, the Convex lens is called a converging lens.
(iii) Calculate the refractive index of the material of the lens shown in the figure.
Solution:
angle of incidence (i) = 90°-54° = 36°
angle of refraction, (r) = 32°
refractive index, (μ) = ?
Now,
From Snell’s law,
\begin{align*} \mu &= \frac{sini}{sinr}\\ &= \frac{sin36 \degree}{sin32 \degree}\\ \therefore \mu &= 1.12 \end{align*}
