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

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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 ​

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    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 )

    is the correct answer.

    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

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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

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

    Answer (Detailed Solution Below)

    Option 3 : 2n−12n+1 Crack with

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    Detailed Solution

    Download Solution PDF

    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.

    Download Solution PDF

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