A rocket of mass $3 \times 10^{6}$ kg takes off from a launching pad and acquires a vertical velocity of 1 km/s at an altitude of 25 km. The kinetic energy is

1.5 \times 10^{10}

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1.5 \times 10^{12}

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1.2 \times 10^{12}

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1.5 \times 10^{8}

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Solution

The correct option is B $1.5 \times 10^{12}$ J
$K E = 0.5 \times m v^{2}$
$\Rightarrow K E = 0.5 \times 3 \times 10^{6} \times (1 \times 1000)^{2} = 1.5 \times 10^{12} J$

Hence, correct answer is option $B$

Note- Here $P E \neq K E$ because the rocket has accelerated i.e. an external force is applied.

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A rocket of mass 3×106 kg takes off from a launching pad and acquires a vertical velocity of 1 km/s at an altitude of 25 km. The kinetic energy is

The correct option is B 1.5×1012 JKE=0.5×mv2⇒KE=0.5×3×106×(1×1000)2=1.5×1012JHence, correct answer is option BNote- Here PE≠KE because the rocket has accelerated i.e. an external force is applied.

A rocket of mass $3 \times 10^{6}$ kg takes off from a launching pad and acquires a vertical velocity of 1 km/s at an altitude of 25 km. The kinetic energy is