{"id":338572,"date":"2023-01-06T10:28:53","date_gmt":"2023-01-06T04:58:53","guid":{"rendered":"https:\/\/infinitylearn.com\/surge\/question\/a-balloon-with-mass-m-is-descending-down-with-an-acceleration-a-where-a-g-how-much-mass-should-be-removed-from-it-so-that-it-starts-moving-up-with-an-acceleration-a-2\/"},"modified":"2025-06-26T17:53:23","modified_gmt":"2025-06-26T12:23:23","slug":"a-balloon-with-mass-m-is-descending-down-with-an-acceleration-a-where-a-g-how-much-mass-should-be-removed-from-it-so-that-it-starts-moving-up-with-an-acceleration-a-2","status":"publish","type":"questions","link":"https:\/\/infinitylearn.com\/surge\/question\/physics\/a-balloon-with-mass-m-is-descending-down-with-an-acceleratio\/","title":{"rendered":"A balloon with mass m is descending down with an acceleration a (where a < g).How much mass should be removed from it so that it starts moving up with an acceleration a?"},"content":{"rendered":"","protected":false},"author":1,"template":"","meta":{"_yoast_wpseo_focuskw":"","_yoast_wpseo_title":"","_yoast_wpseo_metadesc":"A balloon with mass m is descending down with an acceleration a (where a < g).How much mass should be removed from it so that it starts moving up with an acceleration a?","custom_permalink":"question\/physics\/a-balloon-with-mass-m-is-descending-down-with-an-acceleratio\/"},"categories":[],"acf":{"question":"

A balloon with mass m is descending down with an acceleration a (where a < g).How much mass should be removed from it so that it starts moving up with an acceleration a?<\/p>","options":[{"option":"

2<\/mn>ma<\/mi><\/mrow>g<\/mi>+<\/mo>a<\/mi><\/mrow><\/mfrac><\/math><\/p>","correct":true},{"option":"

2<\/mn>ma<\/mi><\/mrow>g<\/mi>-<\/mo>a<\/mi><\/mrow><\/mfrac><\/math><\/p>","correct":false},{"option":"

ma<\/mi>g<\/mi>+<\/mo>a<\/mi><\/mrow><\/mfrac><\/math><\/p>","correct":false},{"option":"

ma<\/mi>g<\/mi>-<\/mo>a<\/mi><\/mrow><\/mfrac><\/math><\/p>","correct":false}],"solution":"

 <\/p>

Forces acting on balloon are its weight and buoyant force (B). The buoyant force will be constant as there is no charge in volume of the balloon.
When balloon is descending down<\/p>

mg - B = ma------------(i)<\/p>

When balloon is moving up<\/p>

B<\/mi>-<\/mo>(<\/mo>m<\/mi>-<\/mo>m<\/mi>0<\/mn><\/msub>)<\/mo>g<\/mi> <\/mo>=<\/mo> <\/mo>(<\/mo>m<\/mi>-<\/mo>m<\/mi>0<\/mn><\/msub>)<\/mo>a<\/mi>(<\/mo>ii<\/mi>)<\/mo><\/math><\/p>
<\/figure>

                                                                      Balloon is moving down              Balloon is moving up<\/p>

Equation (i) + Equation (ii)<\/p>

\u21d2<\/mo> <\/mo>mg<\/mi>-<\/mo>mg<\/mi>+<\/mo>m<\/mi>0<\/mn><\/msub>g<\/mi> <\/mo>=<\/mo>ma<\/mi> <\/mo>+<\/mo> <\/mo>ma<\/mi> <\/mo>=<\/mo> <\/mo>m<\/mi>0<\/mn><\/msub>a<\/mi> <\/mo><\/math><\/p>

\u21d2<\/mo>m<\/mi>0<\/mn><\/msub> <\/mo>=<\/mo> <\/mo>2<\/mn>ma<\/mi><\/mrow>g<\/mi>+<\/mo>a<\/mi><\/mrow><\/mfrac><\/math><\/p>

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