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Preeti Dabral 3 years, 1 month ago
Let the radius and height of the cylinder be r and h, respectively.
Curved surface area of cylinder = 2{tex}\pi{/tex}rh
Area of base = {tex}\pi{/tex}r2
Sum of areas of bases = 2{tex}\pi{/tex}r2
According to the question, 4 {tex}\times{/tex} Curved surface area = 6 {tex}\times{/tex} Sum of areas of bases
4 {tex}\times{/tex} 2{tex}\pi{/tex}rh = 6 {tex}\times{/tex} 2{tex}\pi{/tex}r2
= 8{tex}\pi{/tex}rh = 12{tex}\pi{/tex}r2
= 2h = 3r
{tex}\Rightarrow \quad r=\frac{2}{3} h{/tex}
{tex}\therefore \quad r=\frac{2}{3} \times 12{/tex} = 8cm [{tex}\because{/tex} h = 12 cm, given]
{tex}\therefore{/tex} Curved surface area of the cylinder = 2{tex}\pi{/tex}rh
= {tex}2 \times \frac{22}{7} \times 8 \times 12=\frac{44 \times 8 \times 12}{7}{/tex}
= 603.428 cm2
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Preeti Dabral 3 years, 1 month ago
On 13th April 1919, a large crowd gathered in the enclosed ground of Jallianwala Bagh. Some to protest against the British government’s repressive measures, others to attend the annual Baishakhi Fair. These people were unaware of the imposition of Martial Law in the city. General Dyer, the Commander, blocked the exit points from the Bagh and opened fire upon the innocent citizens. Dyer’s intention was to produce a moral effect and terrorize Satyagrahis. Hundreds of innocent people including women and children were killed and wounded due to this indiscriminate firing by the British soldiers, which ultimately led to nation-wide outrage. Jallianwala Bagh incident was the most brutal incident in the History of India.
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Preeti Dabral 3 years, 1 month ago
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