NEB Grade 11 Physics Model Paper Solution

6. (a) Sketch an electric field pattern around two identical negative point charges shown below.
Ans:

(b) Obtain an equation, in terms of Q and r, for the field strength at point X due to two charges shown in Fig. 6.1.

Solution:

The electric field at X due to the first charge is given by:

E_{1} = \frac{Q}{4 \pi \epsilon _{o}(2r)^{2}}

Also, the electric field due to the second charge is:

E_{2} = \frac{Q}{4 \pi \epsilon _{o}r^{2}}

Since both charges are –ve and their field direction are the same, the net electric field at X is given by:

\begin{align*} E_{net} &= E_{1} + E_{2}\\ \therefore\ E &= \frac{Q}{4 \pi \epsilon _{o}(2r)^{2}}\ +\ \frac{Q}{4 \pi \epsilon _{o}r^{2}} \end{align*}

7. (a) Define the capacitance of a parallel plate capacitor and state one application of it in an electric circuit.
Ans: Capacitance of parallel plate capacitor is defined as the ability of the parallel plate capacitor to store the charge. Mathematically,

C = \frac{\epsilon _{o}A}{d}

Where ϵo is the permittivity of free space, A is the area of the plate of the capacitor and d is the distance between the plates.

(b) Three capacitors each of 1000μF are connected in an electric circuit as shown below.

(i) Identify the type of combination shown in Fig. 7.1, and calculate the effective capacitance of the combination.
Solution:

The fig above can also be drawn as:

From above, it is clear that they are in parallel combinations. The effective capacitance for parallel combination is given by:

\begin{align*} C_{eff} &= C_{1} + C_{2} + C_{3}\\ &= 1000 + 1000 + 1000\\ \therefore\ C_{eff} &= 3000\ \mu F \end{align*}

8. (a) What is it meant by the power of a heater is 2 kW?
Ans: It means that every second 2000 J of electrical energy is converted into thermal energy by the heater.

(b) Calculate the resistance of the above-mentioned heater when it is connected to 220V source.
Solution:
Given,
Power (P) = 2 kW = 2000 W
volt (V) = 220 V
Resistance (R) =?
We know that,

\begin{align*} P&=\frac{V^{2}}{R}\\ Or,\ R &= \frac{(220)^{2}}{2000}\\ \therefore\ R &= 24.2\ \Omega \end{align*}

(c) Suggest what changes must be done to the heater so that it gives more heat. Justify your answer.
Ans: Heat produced in a conductor due to flow of current is given by:

H = I^{2}RT

Since the resistance of the heater remains constant, more heat can be produced by increasing the current and passing current through the wire of the heater for a longer period of time.

Group C

9. A box at rest is accelerated by a rope attached with a motor as shown in the Fig 2.1. The velocity-time graph given below shows the pattern of its motion for 20 s.

(a) If the box is pulled with constant unbalanced force 10N. Show that the initial acceleration of the box is 2.5 ms-2, and calculate its mass.
Solution:
Let us consider the motion of the box from time t=0 to t=2 sec.
From above graph,
V_o = 0m/s
V₂ = 5 m/s

\begin{align*} so,\ a &= \frac{V_{2}-V_{o}}{t_{2}-t_{o}}\\ &= \frac{5-0}{2-0}\\ \therefore\ a&= 2.5\ m/s^{2}\\ Again,\\ F &= m \times a\\ or,\ 10 &= m \times 2.5\\ \therefore\ m &= 4\ kg. \end{align*}

(b) After 2.0 second the box is being pulled by a constant force 12 N. Determine the size of frictional forces acting on the box at this time.
Solution:
Here, F= 12N.
Since frictional force comes into act,

\begin{align*} F_{net} - F_{r} &= m \times a \\ or,\ 12 - F_{r} &= 4 \times 2.5 \\ \therefore\ F_{r} &= 2N. \end{align*}

(c) Determine the distance of the box travels along the ground at 8.0s.
Solution:
To calculate the distance that the box travels along the ground at 8 sec, we have to find the area under V-t graph up to 8 sec.
so at 8 sec,

t = 8\\ \Delta v = 20

Therefore, distance travelled is given by:

\begin{align*} d &= \frac{1}{2} \times \Delta v \times t\\ &= \frac{1}{2} \times 20 \times 8\\ &= 80\ m. \end{align*}

10. A boy is operating a remote-controlled toy car on a horizontal circular track, as shown in Figure. The track has a radius of 1·8 m and the car travels around the track with a constant speed.

(i) Explain why the car is accelerating, even though it is travelling at a constant speed.
Ans: Even though the car is travelling at a constant speed, its velocity is changing at each second of its motion which causes an acceleration in the car i.e. centripetal acceleration.

(ii) The car has a mass of 0·50 kg. The boy now increases the speed of the car to 6·0 m s -1. The total radial friction between the car and the track has a maximum value of 7.0 N. Show by calculation that the car cannot continue to travel in a circular path.
Solution:
Radius of the track (r) = 1.8m
Mass of car (m) = 0.5kg
Speed of the car (v) = 6 m/s
Maximum radial friction (Fr) = 7 N

The necessary centripetal force required by the car to continue its motion in a uniform circular field is provided by the frictional force between the car and the track. So, centripetal force is given by:

\begin{align*} F_{c} &= \frac{mv^{2}}{r}\\ or,&= \frac{0.5 \times 6^{2}}{1.8}\\ \therefore\ F_{c} &= 10\ N \end{align*}

Since, the frictional force between the car and track (7N) is less than the required centripetal force (10N), the car cannot continue to travel in the circular path. It will slip out of the track.

OR

Juno is a NASA orbiter with a mission to survey Jupiter. It is in an elliptical orbit around Jupiter as shown in the figure below.

The gravitational potential at point A in the orbit of Juno is -1·70 × 109 J kg-1.

(a) State what is meant by a gravitational potential at point A is -1·70 × 109 J kg-1.
Ans: It means that 1.7 x 10⁹ J of energy is required by Jupiter to take a unit mass from infinity to a certain point in its gravitational field.

(b) At point B, Juno is 1·69 × 10⁸ m from the centre of Jupiter. If the mass if Jupiter is 1.90 x 10²⁷ kg, calculate the gravitational potential at point B.
Solution:
Gravitational potential at point B is given by:

\begin{align*} V_{B} &= \frac{-GM}{r}\\ &= \frac{6.67 \times 10^{-11} \times 1.9 \times 10^{27}}{1.69 \times 10^{8}}\\ &= -7.5 \times 10^{8}\ J/Kg \end{align*}

(c) The mass of Juno is 1·6 × 10³ kg. Determine the change in gravitational potential energy if Juno moves from Point A to Point B.
Solution:
The gravitation potential energy is given by:

\begin{align*} U = -\frac{GMm}{r} = -V \times m\\ \end{align*}

So, the change in gravitational potential energy in moving Juno from point A to B is given by:

\begin{align*} U_{A} - U_{B} &= -V_{A} \times m -(-V_{B}) \times m\\ &= m(V_{B}-V_{A})\\ &= 1.6 \times 10^{3} \times (-7.5 \times 10^{8} + 1.7 \times 10^{9})\\ \therefore\ U_{A} - U_{B} &= -1.52 \times 10^{12}\ J. \end{align*}