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A block of mass 5 kg moves along the x-direction subject to the force F = (−20x + 10) N, with the value of x in metre. At time t = 0 s, it is at rest at position x=1m. the position and momentum of the block at t=pi/4s are

on June 28, 2025

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To solve the problem, we'll follow these steps: Step 1: Assign Coordinates Let’s place the triangle Δ P Q R Δ PQR in the plane for simplicity: Let P = ( 0 , 0 ) P = ( 0 , 0 ) , Q = ( 1 , 0 ) Q = ( 1 , 0 ) , and R = ( 0 , 1 ) R = ( 0 , 1 ) . Step 2: Find the Coordinates of S S Given the vector equation: S P ‾ + 5 S Q ‾ + 6 S R ‾ = 0 ‾ , SP + 5 SQ + 6 SR = 0 , we express the vectors in terms of coordinates: S P ‾ = P − S = ( 0 − x , 0 − y ) , SP = P − S = ( 0 − x , 0 − y ) , S Q ‾ = Q − S = ( 1 − x , 0 − y ) , SQ = Q − S = ( 1 − x , 0 − y ) , S R ‾ = R − S = ( 0 − x , 1 − y ) . SR = R − S = ( 0 − x , 1 − y ) . Substituting into the equation: ( 0 − x , 0 − y ) + 5 ( 1 − x , 0 − y ) + 6 ( 0 − x , 1 − y ) = ( 0 , 0 ) . ( 0 − x , 0 − y ) + 5 ( 1 − x , 0 − y ) + 6 ( 0 − x , 1 − y ) = ( 0 , 0 ) . This gives the system: − x + 5 ( 1 − x ) + 6 ( − x ) = 0 and − y + 5 ( − y ) + 6 ( 1 − y ) = 0. − x + 5 ( 1 − x ) + 6 ( − x ) = 0 and − y + 5 ( − y ) + 6 ...

Problem Statement: We are given: A parabola with equation y 2 = 4 a x y 2 = 4 a x where a > 0 a > 0 . A point P P on this parabola. The normal to the parabola at P P meets the x-axis at point Q Q . The area of triangle P F Q PFQ is 120, where F F is the focus of the parabola. The slope m m of the normal and a a are both positive integers. We need to find the correct pair ( a , m ) ( a , m ) from the given options. Step 1: Understand the Parabola and Focus The standard parabola y 2 = 4 a x y 2 = 4 a x has: Vertex at ( 0 , 0 ) ( 0 , 0 ) . Focus F F at ( a , 0 ) ( a , 0 ) . Step 2: Parametrize Point P P on the Parabola Let P P be a point on the parabola. We can parametrize P P as: P = ( a t 2 , 2 a t ) , P = ( a t 2 , 2 a t ) , where t t is a parameter. Step 3: Find the Equation of the No...

Math JEE Adv. 2025 paper 1

To solve the problem, we'll count the number of seven-digit numbers formed using digits {0, 1, 2} where at least one of the digits 0 or 1 appears exactly twice. Here's the step-by-step solution: Step 1: Total Possible Numbers Each of the 7 digits can be 0, 1, or 2. The first digit cannot be 0 (to ensure it's a seven-digit number). Total numbers: 2 × 3 6 = 2 × 729 = 1458 2 × 3 6 = 2 × 729 = 1458 . Step 2: Numbers Where 0 Appears Exactly Twice Choose positions for 0: ( 6 2 ) ( 2 6 ) (since the first digit cannot be 0). Fill the remaining 5 digits: Each can be 1 or 2 (since we're fixing exactly two 0s). Count: ( 6 2 ) × 2 5 = 15 × 32 = 480 ( 2 6 ) × 2 5 = 15 × 32 = 480 . Step 3: Numbers Where 1 Appears Exactly Twice Choose positions for 1: ( 7 2 ) ( 2 7 ) . Fill the remaining 5 digits: Each can be 0 or 2 (but the first digit cannot be 0). If the first digit is 1: ( 6 1 ) × 2 5 = 6 × 32 = 192 ( 1 6 ) × 2 5 = 6 × 32 = 192 . I...

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