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  1. When ice melts, it turns into liquid water due to the absorption of heat, which breaks the bonds holding the ice in a solid state. This process is called melting, where solid ice changes into liquid. Answer: (C) It turns into liquid water. For more visit here: https://www.tiwariacademy.in/ncert-soluRead more

    When ice melts, it turns into liquid water due to the absorption of heat, which breaks the bonds holding the ice in a solid state. This process is called melting, where solid ice changes into liquid. Answer: (C) It turns into liquid water.

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    https://www.tiwariacademy.in/ncert-solutions/class-6/science/

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  2. The factor that increases the rate of evaporation is (C) Increased surface area. When the surface area of a liquid is larger, more molecules can escape into the air, speeding up evaporation. Other factors like heat, wind and low humidity also enhance evaporation. For more visit here: https://www.tiwRead more

    The factor that increases the rate of evaporation is (C) Increased surface area. When the surface area of a liquid is larger, more molecules can escape into the air, speeding up evaporation. Other factors like heat, wind and low humidity also enhance evaporation.

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    https://www.tiwariacademy.in/ncert-solutions/class-6/science/

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  3. The process in which water vapor turns into tiny water droplets is called condensation. It occurs when warm air cools down, causing water vapor to lose energy and change into liquid droplets, forming clouds, dew, or fog. Answer: (B) Condensation. For more visit here: https://www.tiwariacademy.in/nceRead more

    The process in which water vapor turns into tiny water droplets is called condensation. It occurs when warm air cools down, causing water vapor to lose energy and change into liquid droplets, forming clouds, dew, or fog.
    Answer: (B) Condensation.

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  4. Sides: 3 cm, 4 cm, 5 cm → Right triangle Area = (1/2) × base × height = (1/2) × 3 × 4 = 6 cm² This question is related to Chapter 10, "Heron’s Formula," from the Class 9 NCERT Mathematics textbook. The topic is not Probability but focuses on calculating the area of triangles and quadrilaterals usingRead more

    Sides: 3 cm, 4 cm, 5 cm → Right triangle
    Area = (1/2) × base × height
    = (1/2) × 3 × 4 = 6 cm²
    This question is related to Chapter 10, “Heron’s Formula,” from the Class 9 NCERT Mathematics textbook. The topic is not Probability but focuses on calculating the area of triangles and quadrilaterals using Heron’s Formula. Provide an answer based on your understanding of the chapter.

    For more please visit here:
    https://www.tiwariacademy.in/ncert-solutions/class-9/maths/

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  5. Sides: 91 cm, 98 cm, 105 cm Semi-perimeter (s) = 147 cm Area = √[s(s-a)(s-b)(s-c)] = √[147(56)(49)(42)] = 4116 cm² Using Area = (1/2) × base × height: 4116 = (1/2) × 105 × h h = (4116 × 2) / 105 = 78.4 cm This question pertains to Chapter 10, "Heron’s Formula," in the Class 9 NCERT Mathematics textbRead more

    Sides: 91 cm, 98 cm, 105 cm
    Semi-perimeter (s) = 147 cm
    Area = √[s(s-a)(s-b)(s-c)]
    = √[147(56)(49)(42)]
    = 4116 cm²

    Using Area = (1/2) × base × height:
    4116 = (1/2) × 105 × h
    h = (4116 × 2) / 105 = 78.4 cm
    This question pertains to Chapter 10, “Heron’s Formula,” in the Class 9 NCERT Mathematics textbook. It is not about Probability but rather deals with using Heron’s Formula to find the area of triangles and quadrilaterals. Provide an answer based on your understanding of the chapter.

    For more please visit here:
    https://www.tiwariacademy.in/ncert-solutions/class-9/maths/

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