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20 g : 100 g = 1 metre : 500 cm...

20 g : 100 g = 1 metre : 500 cm

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Knowledge Check

  • The solubility of common salt is 36.0 g in 100 g of water at 20^(@) C . If systems I , II and III contains 20.0 , 18.0 "and" 15.0 of the salt added to 50.0 g of water in each case , the vapour pressures would be in the order :

    A
    `I lt II lt III `
    B
    `I gt II gt III`
    C
    `I = II gt III`
    D
    `I = II lt III`

  • Directions for questions : Select the correct answer from the given options. The cost price C is given by the formula. C = (100S)/( 100 + g) , where S = selling g price and g= gain in %. Make S the subject of the formula. Find S, if C = Rs. 800 and g = 20 . The following steps are involved in solving the above problem. Arrange them in sequential order. A rArr 100 S = C (100 +g) B S = ((100 + g)C)/( 100) C Given C = (100 S)/( 100 +g) D therefore S = ((100 + 20 ) xx 800)/( 100) = Rs. 960

    A
    ABCD
    B
    BCAD
    C
    CADB
    D
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  • Three masses are placed on the x-axis : 300 g at origin , 500 g at x = 40 cm and 400 g at x = 70 cm . The distance of the centre of mass from the origin is

    A
    40 cm
    B
    45 cm
    C
    50 cm
    D
    30 cm

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    The solubility of common salt is 36.0 g in 100 g of water at 20 ^@C . If systems, I, II and III contain 40.0, 36,0 and 20.0 g of the salt added to 100.0 g of water in each case, the vapour pressures would be in the order

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