JEE Mains

    24 Jan 2023 - Shift 1

    Physics

    QuestionA travelling wave is described by the equation

    y(x,t)=[.05sin(8x−4t)]y(x,t) = [.05sin(8x-4t)]

    The velocity of the wave is : [ all the quantities are in SI unit ]

    (A)0.5 ms-1 (B) 8 ms-1 (C) 2 ms -1 (D) 4 ms-1

    AnswerVelocity of Sine Wave=Angular FrequencyWave Number\text{Velocity of Sine Wave} = \frac{\text{Angular Frequency}}{\text{Wave Number}}

    Angular Frequency = 4 and Wave number is 8 So the answer is .5 ms−1m {s}^{-1}

    Chemistry

    Q1
    AnswerWe know that square root of frequency of x-ray is directly proportional to the atomic number! So square root of v = v raised to n. That's why value of n is 1/2.

    Maths

    Questionlim⁡t→0(11sin⁡2(t)+21sin⁡2(t)+...+n1sin⁡2(t))sin⁡2(t)\lim_{t \to 0} (1^\frac{1}{\sin^2(t)} + 2^\frac{1}{\sin^2(t)} + ... + n^\frac{1}{\sin^2(t)}) ^ {\sin^2(t)}  is equal to ...
    AnswerRemember that  a∞=0 if 0 < a < 1 a^\infty = 0 \text { if 0 < a < 1 }

    So  lim⁡t→0(11sin⁡2(t)+21sin⁡2(t)+...+n1sin⁡2(t))sin⁡2(t)\lim_{t \to 0} (1^\frac{1}{\sin^2(t)} + 2^\frac{1}{\sin^2(t)} + ... + n^\frac{1}{\sin^2(t)}) ^ {\sin^2(t)}

    is equal to  lim⁡t→0(1cos⁡2(t)+2cos⁡2(t)+...+ncos⁡2(t))sin⁡2(t)\lim_{t \to 0} (1^{\cos^2(t)} + 2^{\cos^2(t)} + ... + n^{\cos^2(t)}) ^ {\sin^2(t)}

    is equal to  lim⁡t→0((ncos⁡2(t))sin⁡2(t))((1n)cos⁡2(t)+(2n)cos⁡2(t)+...+(nn−1)cos⁡2(t)+1)sin⁡2(t)\lim_{t \to 0} ( (n^{\cos^2(t)}) ^ {\sin^2(t)}) ((\frac{1}{n})^{\cos^2(t)} + (\frac{2}{n})^{\cos^2(t)} + ... + (\frac{n}{n-1})^{\cos^2(t)} + 1) ^ {\sin^2(t)}

    is equal to nn

    24 Jan 2023 - Shift 2