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  1. When a body is moving in a circular path with constant speed, we have to look into the consequences of circular motion. 1. Work done will be zero: - True. Since the force (centripetal force) is always perpendicular to the direction of motion, no work is done on the body. 2. Acceleration will be zeroRead more

    When a body is moving in a circular path with constant speed, we have to look into the consequences of circular motion.

    1. Work done will be zero:
    – True. Since the force (centripetal force) is always perpendicular to the direction of motion, no work is done on the body.

    2. Acceleration will be zero:
    – False. The velocity has a constant speed but is changing direction all the time, and hence there is a centripetal acceleration directed toward the center of the circular path.

    3. No force acts on the body:
    – False. A net force (centripetal force) is necessary to keep the body moving in a circular path and is directed towards the center.

    4. Its velocity remains constant:
    – False. True, the velocity magnitude is unchanged, but the velocity is a vector: its direction must change; it is not a constant.

    Final Answer:
    Work done will be zero.

    Click here :
    https://www.tiwariacademy.com/ncert-solutions/class-11/physics/chapter-5/

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  2. In the photon nature of light, the intensity of light is determined by the number of photons incident per unit area per second. A higher photon count corresponds to greater energy delivery, thereby increasing the light's intensity. For more visit here: https://www.tiwariacademy.com/ncert-solutions/cRead more

    In the photon nature of light, the intensity of light is determined by the number of photons incident per unit area per second. A higher photon count corresponds to greater energy delivery, thereby increasing the light’s intensity.

    For more visit here:
    https://www.tiwariacademy.com/ncert-solutions/class-12/physics/chapter-11/

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  3. Incorporating Yoga into a daily routine provides numerous benefits, including improved physical strength, flexibility, and posture. It enhances mental clarity, focus, and emotional stability while reducing stress and anxiety. Regular Yoga practice promotes mindfulness, boosts energy levels, and fostRead more

    Incorporating Yoga into a daily routine provides numerous benefits, including improved physical strength, flexibility, and posture. It enhances mental clarity, focus, and emotional stability while reducing stress and anxiety. Regular Yoga practice promotes mindfulness, boosts energy levels, and fosters inner peace. It also encourages a positive outlook and self-discipline, enabling individuals to manage their responsibilities more effectively. Ultimately, Yoga supports a harmonious balance between body and mind, contributing to a healthier, happier lifestyle.

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  4. Santosha, the practice of contentment in Niyama, teaches individuals to appreciate what they have rather than focusing on what they lack. This mindset helps reduce desires and stress, creating a sense of fulfillment and gratitude. By accepting life’s challenges gracefully, Santosha fosters resiliencRead more

    Santosha, the practice of contentment in Niyama, teaches individuals to appreciate what they have rather than focusing on what they lack. This mindset helps reduce desires and stress, creating a sense of fulfillment and gratitude. By accepting life’s challenges gracefully, Santosha fosters resilience and inner calm. Practicing contentment enhances relationships and overall well-being, as one learns to cherish simple joys and remain happy irrespective of external circumstances, leading to sustainable personal happiness.

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  5. To find the ratio of the potential energies of two springs with different spring constants when stretched by the same force, we can use the formula for elastic potential energy: U = (1/2) k x² Where: - U is the potential energy - k is the spring constant - x is the extension of the spring Step 1: ReRead more

    To find the ratio of the potential energies of two springs with different spring constants when stretched by the same force, we can use the formula for elastic potential energy:

    U = (1/2) k x²

    Where:
    – U is the potential energy
    – k is the spring constant
    – x is the extension of the spring

    Step 1: Relationship between force and extension

    According to Hooke’s Law, the force applied to a spring is related to the spring constant and the extension by:

    F = kx

    Rearranging gives us:

    x = F/k

    Step 2: Calculate potential energy for both springs

    For spring 1 (k₁ = 1500 N/m):
    U₁ = (1/2) k₁ x₁²
    Substituting for x₁:
    U₁ = (1/2) k₁ (F/k₁)²
    U₁ = (1/2) k₁ (F²/k₁²)
    U₁ = (1/2) (F²/k₁)

    For spring 2 (k₂ = 3000 N/m):
    U₂ = (1/2) k₂ x₂²
    Substituting for x₂:
    U₂ = (1/2) k₂ (F/k₂)²
    U₂ = (1/2) k₂ (F²/k₂²)
    U₂ = (1/2) (F²/k₂)

    Step 3: Find the ratio of potential energies

    Now we can find the ratio of U₁ to U₂:

    U₁ / U₂ = (F² / (2k₁)) / (F² / (2k₂))
    U₁ / U₂ = k₂ / k₁
    U₁ / U₂ = 3000 / 1500
    U₁ / U₂ = 2

    Final Answer:
    The potential energies of the two springs will be in the ratio of 2 : 1.

    Click for more:
    https://www.tiwariacademy.com/ncert-solutions/class-11/physics/chapter-5/

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