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A solid weight 0.09N in air and0.06N in water what is R.d of the solid |
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Answer» Given, WEIGHT in air (W₁) = 0.09 N Weight in water (W₂) = 0.06 N R.D = W₁/(W₁-W₂) = 0.09 / (0.09-0.06) = 0.09/0.03 = 3 The Relative DENSITY (R.D) is 3. I hope this HELPS. |
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