Question

The length of rod is 100 m. If
length of this rod is measured by
the observer moving parallel to its
length is 51 m, find the speed of
observer.
(A) 0.86 C
(B) 0.80 C
(C) 0.92 C
(D) 0.96 C​

Answer 1

Answer:

Explanation:

Given Rest length L. = 1m Length of the moving rod L = Lo√1- () ² where c = 3 × 108 m/s (a): v= 3 x 105 m/s

⇒ L₁=1 x

3 x 105 3 × 108)² = 0.99999 m

(b) v = 3 x 106 m/s

⇒ L₁=1 x 0.99995 m 1 3 × 106, 5) ² 3 x 108 =

(c): v= 3 x 10¹ m/s 1 3 x 107 ) ² 3 x 108 - ⇒ Lb=1 x 0.99499 m

Answer 2

The speed of the observer will be 0.86c. (option A)

Given,

Length of a rod = 100 m.

Length measured by the observer = 51 m.

To find,

Speed of the observer.

Solution,

We know, from Einstein's theory of special relativity that, when an object is moving relative to an observer who is at rest, the object in motion appears shorter to the observer. It is called length contraction.

It can be observed that it will be the same, whether the object moves relative to the stationary observer, or the observer moves relative to the stationary object.

For an object having the actual length $L_o$, moving with velocity v relative to the stationary observer, the measured or observed length (L), will be given by

$L=L_o\sqrt{1-\frac{v^2}{c^2} } \hfill ...(1)$

where c is the speed of light in a vacuum.

Substituting the respective given values in (1), we get,

$51=100 \times \sqrt{1-\frac{v^2}{c^2} }$

Rearranging and simplifying,

$\sqrt{1-\frac{v^2}{c^2} }=\frac{51}{100}$

Squaring both sides, we get,

$1-\frac{v^2}{c^2} =(0.51)^2$

$\implies \frac{v^2}{c^2} =1-0.2601$

$\implies v^2 =0.7399 \times c^2$

$\implies v =\sqrt{0.7399} \times c$

⇒ v = 0.8602c

⇒ speed of the observer = v = 0.86c.

Therefore, the speed of the observer will be 0.86c. (option A)

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