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# a small block slides down on a smooth inclined plane starting from rest

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### Mohammed

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## A small block slides without friction down an inclined plane starting from rest. Let Sn be the distance travelled from time t = ( n

Click here👆to get an answer to your question ✍️ A small block slides without friction down an inclined plane starting from rest. Let Sn be the distance travelled from time t = ( n - 1 ) to t = n . The SnSn + 1 is.

A small block slides without friction down an inclined plane starting from rest. Let S

Question n ​

be the distance travelled from time t=(n−1)tot=n. The

S n+1 ​ S n ​ ​ is.

A

2n (2n−1) ​

B

(2n−1) (2n+1) ​

C

(2n+1) (2n−1) ​

D

(2n+1) (2n) ​ Medium Open in App

Updated on : 2022-09-05

Solution Verified by Toppr

Correct option is C)

For distance S n ​

travelled from (n−1) to n

S n ​ =0[n−n−1]+ 2 1 ​ a[n 2 −(n−1) 2 ] S n ​ = 2 1 ​ a[n 2 −(n 2 +1−2n)] S n ​ = 2 a ​ (2n−1) Again , S n+1 ​ = 2 1 ​ a[(n+1) 2 −n 2 ] = 2 a ​ [2n+1] S n+1 ​ S n ​ ​ = 2 a ​ (2n+1) 2 a ​ (2n−1) ​ = 2n+1 2n−1 ​

483 81

स्रोत : www.toppr.com

## A small block slides down on a smooth inclined plane, starting from rest at time t = 0. Let Sn be the distance travelled by the block in the interval, t = n 1 to t = n, then, the ratio SnSn+1 is

A small block slides down on a smooth inclined plane, starting from rest at time t = 0. Let Sn be the distance travelled by the block in the interval, t = n 1 to t = n, then, the ratio SnSn+1 is

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A small block slides down on a smooth inclined plane, starting from rest at time t = 0. Let Sn be the distance travelled by the block in the interval, t = n 1 to t = n, then, the ratio SnSn+1 is

Question

A small block slides down on a smooth inclined plane, starting from rest at time

t = 0. Let S n

be the distance travelled by the block in the interval,

t = n − 1 to t = n , then, the ratio S n S n + 1 is A 2 n 2 n − 1 B 2 n − 1 2 n C 2 n − 1 2 n + 1 D 2 n + 1 2 n − 1 Open in App Solution

The correct option is C

2 n − 1 2 n + 1 Suppose, θ

is inclination of inclined plane.

So, the acceleration along inclined plane

a = g sin θ Initial speed, u = 0 .

By equation of uniformly accelerated motion, distance travelled by object during

n t h second, S n = u + a 2 ( 2 n − 1 ) S n = 0 + g sin θ 2 ( 2 n − 1 ) = g sin θ 2 ( 2 n − 1 ) . . . ( i )

Distance travelled during

( n + 1 ) t h second. S n + 1 = 0 + g sin θ 2 [ 2 ( n + 1 ) − 1 ] = g sin θ 2 ( 2 n + 1 ) . . . ( i i ) Using ( 1 ) and ( 2 ) , S n S n + 1 = ( 2 n − 1 ) ( 2 n + 1 ) Hence, ( c )

Suggest Corrections 48

SIMILAR QUESTIONS

Q. A small block slides down on a smooth inclined plane, starting from rest at time

t = 0. Let S n

be the distance travelled by the block in the interval,

t = n − 1 to t = n , then, the ratio S n S n + 1 is

Q. A block is moving down a smooth inclined plane starting from rest at time

t = 0 . Let S n

be the distance travelled by the block in the interval

t = n − 1 to t = n . The ratio S n S n + 1 is

Q. A smallA block slides a smooth inclined plane, starting from rest at time t = 0.Let Sn be the distance travelled by th block in the interval t=n-1 to t=n.Then the Ratioos Sn/Sn-1Q. a small block slides down a smooth inclined plane,starting from rest at time t=0.let sn be the distance travelled by the block in the interval t=n-1 to t=n.Then,the ratio sn/sn+1 is: (a)2n-1/2n (b)2n-1/2n+1 (c)2n+1/2n-1 (d)2n/2n-1Q.

A small block slides without friction down the inclined plane starting from rest . let Sn be the distance travelled from t = n-1 to t= n. Then Sn/Sn + 1 is ?

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स्रोत : byjus.com

## [Solved] A small block slides down on a smooth inclined plane, starti

CONCEPT: According to the graphical second equation of motion for an nth second we have; Sn = u + $$\frac{a}{2}$$ (2n - 1) ---(1) where&n

Home Physics Motion in a Straight Line Kinematic equations for uniformly accelerated motion

## Question

A small block slides down on a smooth inclined plane, starting from rest at time t=0. Let Sn be the distance travelled by the block in the interval

t=n−1 to t=n. Then, the ratio

n n1 SnSn+1 is : 2n2n−1 2n−12n 2n−12n+1 2n+12n−1

Option 3 : 2n−12n+1 Crack with

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## Detailed Solution

CONCEPT:

According to the graphical second equation of motion for an nth second we have;

Sn = u + a2 (2n - 1) ---(1)

where Sn = distance traveled by an object during the nth second.

u is the initial velocity.

a is the acceleration.

CALCULATION:

Suppose θ is the inclination of inclined plane acceleration along the inclined plane is written as;

a = g sin θ

By using equation (1) of uniformly accelerated motion we have;

Sn = u + a2 (2n - 1)

Putting the value of acceleration 'a' and initial velocity, u is zero, we have;

Sn=0+gsin⁡θ2(2n−1) ⇒ Sn=gsin⁡θ2(2n−1) -----(2)

Distance traveled during (n + 1)th second.

Sn+1 = 0 + gsin⁡θ2[2(n+1)−1] Sn+1=gsin⁡θ2(2n−1) -----(3)

Dividing equation (2) and (3) we have;

SnSn+1=(2n−1)(2n+1)

Hence, option 3) is the correct answer.

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Mohammed 7 day ago

Guys, does anyone know the answer?