Question

20. A train passes an electric post in 10 second and a bridge of length 2 km in 110 second. The speed of the
engine is
(A) 18 kmph
(B) 36 kmph
(C) 72 kmph
(D) 90 kmph​

Answer 1

Answer :

Let x be the length of the train and v be the velocity.

Case 1 :

Train takes 10 s to pàss an electric post.

We know, Velocity = Distance/Time —(1)

Hence, v = x/10 m/s —(2)

Case 2 :

Train takes 110 s to pàss 2 km (2000 m) length bridge.

Hence, by using (1), v = (x+2000)/110 m/s —(3)

Therefore, Velocity = v = (1) = (2)

⇒ x/10 = (x+2000)/110

⇒ 11x = (x+2000)

⇒ 11x-x = 2000

⇒ 10x = 2000

⇒ x = 200 m —(4)

So, the length of the train = x = 200 m

Now, By using (2) and (4),

⇒ v = 200/10

⇒ 20 m/s

Converting :

⇒ v = 20 × 18/5

⇒ 4 × 18

⇒ 72 km/h

So, the velocity of the train = v = 72 kmph – Option (C)

Answer 2

$\underline{\underline{\textsf{\textbf{\blue{\quad Correct\:Question\::\quad}}}}}$

A train passes an electric post in 10 seconds and a bridge of length 2km in 110 seconds. The speed of the train is :

(A) 18 km/h

(B) 36 km/h

(C) 72 km/h

(D) 90 km/h

$\underline{\underline{\textsf{\textbf{\blue{\quad Answer\::\quad}}}}}$

• Option (C) 72 km/h is correct.

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$\underline{\underline{\textsf{\textbf{\blue{\quad Explanation\::\quad}}}}}$

• Let l be the length of the train and x be the velocity of train.

$\underline{\underline{\textsf{\textbf{\blue{\quad In\:first\:case\::\quad}}}}}$

• A train passes an electric post in 10 seconds. So, x = l/10 m/s ————[1]

$\underline{\underline{\textsf{\textbf{\blue{\quad\:In\:second\:case\::\quad}}}}}$

• A train passes a bridge of length 2 km(2000 m) in 110 seconds. So, x = (l + 2000)/110 m/s. ————[2]

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• In both cases velocity(x) of train is same. Therefore, [1] = [2]

ㅤㅤㅤㅤㅤㅤ━━━━━━━━━━

$\dashrightarrow\qquad\sf \dfrac{l}{10} = \dfrac{l + 2000}{110}$

$\dashrightarrow\qquad\sf \dfrac{l}{10}\:\times\:110 = l + 2000$

$\dashrightarrow\qquad\sf \dfrac{l}{1\cancel{0}}\:\times\:11\cancel{0} = l + 2000$

$\dashrightarrow\qquad\sf 11l = l + 2000$

$\dashrightarrow\qquad\sf 11l - l = 2000$

$\dashrightarrow\qquad\sf 10l = 2000$

$\dashrightarrow\qquad\sf l = \dfrac{2000}{10}$

$\dashrightarrow\qquad\sf l = \dfrac{200\cancel{0}}{1\cancel{0}}$

$\dashrightarrow\qquad{\boxed{\frak{\pink{l = 200\:m}}}}\:\purple\bigstar$

∴ Length of the train = l = 200 m

★ Putting value of l in [1] :

$\dashrightarrow\qquad\sf x = \dfrac{200}{10}$

$\dashrightarrow\qquad\sf x = \dfrac{20\cancel{0}}{1\cancel{0}}$

$\dashrightarrow\qquad{\boxed{\frak{\pink{x = 20\:m/s}}}}\:\purple\bigstar$

★ Velocity in km/h :

$\dashrightarrow\qquad\sf x = 20\:\times\:\dfrac{18}{5}$

$\dashrightarrow\qquad\sf x = \cancel{20}\:\times\:\dfrac{18}{\cancel{5}}$

$\dashrightarrow\qquad\sf x = 4 \:\times\:18$

$\dashrightarrow\qquad{\boxed{\frak{\pink{x = 72\:km/h}}}}\:\purple\bigstar$

∴ Velocity of the train = x = 72 km/h

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$\therefore\:{\underline{\sf{Hence,\:option\:\bf{(C)\:72\:km/h}\:\sf{is\:correct}.}}}$