Question

1. which of the following expressions represents escape velocity?

a. $\frac{2gm}{r^2}$
b. $\sqrt{\frac{2gm}{r^2}}$
c. $\sqrt{\frac{2gm}{r}}$
d. $\frac{2gm^2}{r^2}$

1. which of the following observations occurs when a glass rod is rubbed with silk?

a. electrons from the glass are transferred to the silk.
b. protons from the glass are transferred to the silk.
c. electrons from the silk are transferred to the glass.
d. protons from the silk are transferred to the glass.

1. two point charges of 1.0 μc and 1.5 μc are 5.0 cm apart. calculate the magnitude of the force between them. ($\frac{1}{4\pi\epsilon_0} = 9.0 \times 10^9 \\, \text{n m}^2 \text{c}^{-2}$)

a. 2.7 n
b. 5.4 n
c. 6.7 n
d. 7.0 n

1. two parallel plates separated by 15.0 mm have a potential difference of 2.0 kv between them. calculate the electric field intensity between the plates.

a. $1.3 \times 10^3 \\, \text{n c}^{-1}$
b. $3.0 \times 10^4 \\, \text{n c}^{-1}$
c. $7.5 \times 10^3 \\, \text{n c}^{-1}$
d. $4.5 \times 10^4 \\, \text{n c}^{-1}$

1. circuit diagram: 2ω, 6v, e=4v, 6ω. nodes d, c, a, b. currents $i_1$, $i_2$

using kirchhoff’s first law for electrical circuits, the current in the 6 ω resistor in the diagram above will be
a. $i_1 + i_2$
b. $i_1 - i_2$
c. $2i_1$
d. $i_2$

1. four lamps labelled p(110 v, 100 w), q(240 v, 60 w), r(250 v, 40 w) and s(300 v, 25 w) are switched on for the same duration. which of the lamps consumes the highest amount of electrical energy?

a. p
b. q
c. r
d. s

Explanation:

Question 37

Step1: Recall escape velocity formula

The formula for escape velocity \( v_e \) is derived from equating gravitational potential energy to kinetic energy. The formula is \( v_e = \sqrt{\frac{2GM}{R}} \), where \( G \) is the gravitational constant, \( M \) is the mass of the celestial body, and \( R \) is its radius.

Step2: Match with options

Looking at the options, option C is \( \sqrt{\frac{2GM}{R}} \), which matches the escape velocity formula.

When a glass rod is rubbed with silk, glass has a tendency to lose electrons (since it is less electronegative than silk). So electrons from the glass are transferred to the silk. Protons are bound in the nucleus and do not transfer during charging by friction.

Step1: Recall Coulomb's Law

Coulomb's Law is \( F = \frac{1}{4\pi\epsilon_0} \frac{q_1 q_2}{r^2} \), where \( \frac{1}{4\pi\epsilon_0}=9.0\times 10^{9}\ Nm^2C^{-2} \), \( q_1 = 1.0\ \mu C=1.0\times 10^{-6}\ C \), \( q_2 = 1.5\ \mu C = 1.5\times 10^{-6}\ C \), and \( r = 5.0\ cm=0.05\ m \).

Step2: Substitute values into the formula

\[

$$\begin{align*} F&=9.0\times 10^{9}\times\frac{(1.0\times 10^{-6})\times(1.5\times 10^{-6})}{(0.05)^2}\\ &=9.0\times 10^{9}\times\frac{1.5\times 10^{-12}}{0.0025}\\ &=9.0\times 10^{9}\times6\times 10^{-10}\\ &= 5.4\ N \end{align*}$$

\]

Answer:

C. \( \sqrt{\frac{2GM}{R}} \)

Question 38