Question

please solve this question... ​

Answer 1

Answer:

• The Tension (T) in the thread is 0.889 mN

Given:

1. Mass of bob (M) = 80 milligrams.
2. Charge (q) = 2 × 10⁻⁸ C.
3. Electric field (E) = 2 × 10⁴ V/m

Explanation:

$\rule{300}{1.5}$

Here, as we can see that tension is making some angle i.e. θ with the mean position of pendulum, so we need to take the components of tension over here, i.e. T sinθ and T cosθ. we can also clearly see that sine component of tension balances the coulombic force, and cosine component balances the weight of the bob. So,

$\longrightarrow\sf T\sin\theta=F_{c}\quad\dots\dots\sf (1)\\\\\\\longrightarrow\sf T\cos\theta=W\quad\dots\dots\sf (2)$

• Dividing equation (1) by equation (2)

$\longrightarrow\sf \dfrac{T\sin\theta}{T\cos\theta}=\dfrac{F_{c}}{W}\\\\\\\\\longrightarrow\sf \dfrac{\sin\theta}{\cos\theta}=\dfrac{F_{c}}{W}\\\\\\\\\longrightarrow\sf \tan\theta =\dfrac{F_{c}}{W}\\\\\\\\\longrightarrow\sf \tan\theta =\dfrac{qE}{Mg}$

• Substituting the values, {80 milligrams = 80 × 10⁻⁶ Kg}

$\longrightarrow\sf \tan\theta =\dfrac{2\times 10^{-8}\times 2\times 10^{4}}{80\times 10^{-6}\times 10}\\\\\\\\\longrightarrow\sf \tan\theta=\dfrac{4\times 10^{-4}}{800\times 10^{-6}}\\\\\\\\\longrightarrow\sf \tan\theta=\dfrac{4\times 10^{-4}}{8\times 10^{-4}}\\\\\\\\\longrightarrow\sf \tan\theta=\dfrac{4}{8}\\\\\\\\\longrightarrow\sf \tan\theta=\dfrac{1}{2}\\\\\\\\\longrightarrow\sf \theta=\tan^{-1}\Bigg[\dfrac{1}{2}\Bigg]\\\\\\\\\longrightarrow\boxed{\sf \theta\approx 27^{\circ}}$

∴ We got the value of inclination of the thread.

$\rule{300}{1.5}$

$\rule{300}{1.5}$

Now, substituting the value of theta in equation (1)

$\longrightarrow\sf T\sin\theta=F_{c}\quad\dots\dots\sf (1)\\\\\\\\\longrightarrow\sf T\sin 27^{\circ}=qE=4\times10^{-4}\\\\\\\\\longrightarrow\sf T\times 0.45=4\times10^{-4}\\\\\\\\\longrightarrow\sf T=\dfrac{4\times10^{-4}}{0.45}\\\\\\\\\longrightarrow\sf T=8.89\times10^{-4}\\\\\\\\\longrightarrow\sf T=0.889\times10^{-3}\\\\\\\\\longrightarrow\large{\underline{\boxed{\red{\sf T=0.889\;mN}}}}$

∴ The Tension (T) in the thread is 0.889 mN.

$\rule{300}{1.5}$

Answer 2

Answer:

0.899 is the correct answer

Explanation:

hopes it helps you