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When the " wedge and scale " experiment is performed at the rquator , we get $ m = M \left({ y_e \over x_e } \right) $. If the same experiment is perfoormed at the poles , then what is the wright equation ?

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Explanation

Where we take moments of force at equilibrium , the terms of g cancel on both sides $ mg.x_e = Mg.y_e $ $ m = M ( {y_e \over x_e } ) $ The mass is independent of the gravity at that point ; so mass at the poles will be same as mass at the equator

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A force $ 2 \hat i + 3 \hat j $ acts about an axis at a position vector $ ( \hat j + \hat k )$ from the axis, then what is the torque due to the force about the axis ?

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Explanation

$| \vec \tau | = | \vec r \times \vec F | $

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In the experiment of balancing moments, suppose the fulcrum is at the 60 cm mark, and a known mass of 2 kg is used onthe longer arm. The greatest mass of mwhich can be balanced against 2 kg such that the minimum distance of either of the masses from the fulcrum is atleast 10 cm. (Neglect mass of metre scale.) What will be the value of m ?

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Explanation

$ m _{max} = ( 2 kg ) ( {y_{max} \over x_{min}} ) = 12kg $

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The wedge is kept below the 60 cm mark on the meter scale. Known masses of 1 kg and 2 kg are hung at the 20 cmand 30 cm mark respectively. Where will a 4 kg mass be hung on the meter scale to balance it ? (Neglect mass of meter scale.)

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Explanation

The anit -clockwise moments due to 1 kg and 2 kg are = (2 kg wt) (60-20) cm + (2kg wt)(60-30) cm = (1 kg wt) (40)cm + (2kg wt ) (30)cm = 100 kg wt cm. The clock wise moment due to $ 4 kg = 4 kg. \omega t \times x cm $ $ \Rightarrow 100 = 4x or x = 25 cm $ So the 4 kg mass must be hung at (60 cm + x) = (60 cm + 25 cm) = 85 cm mark to balance the scale

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When a metre scale is balanced above a wedge, 1 kg mass is hung at 10 cm mark and a 2 kg mass is hang at the 85 cmmark. To which mark on the meter scale, the fulcrum be shifted (Neglect mass of meter scale) to balance the scale ?

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Explanation

Balancing moment $ 1x =2 (75 - x) \Rightarrow 3x = 150 or x = 50 cm$ Fulcrum is at 10 cm + 50 cm = 60 cm mark.

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Arrange rubber, steel and glass in the order of decreasing elasticity.

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The following wiresA, B, C, D, are made of the same material with diff. length and diameter. Which of these will have the largest extension, when the same tension is applied ?

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Explanation

$ y = { F.L \over Al } \Rightarrow l = { F.L \over ( {\pi d^2 \over 4} ) y }so \;l \alpha \;( {L \over d^2 }) $ $ l _A \gt l_B \gt l_C \gt l_D $

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The compressibility of a substance equals..........

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Explanation

$ {1 \over k } = { \triangle V \over PV } $

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Two rods of different materials having coefficients of thermal expansions $ \alpha_1 , \alpha_2$ and Young's moduli $y_1, y_2$ respectively are fixed between two rigid walls. The rods are heated such that they undergo the same increase in temperature. There is no bending of the rod. if $ \alpha_1 : \alpha_2 = 2:3 $ thermal stresses developed in the rod are equal provided $y_1 : y_2$ equals.

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Explanation

$ \left( F \over A \right) = y \left({ l \over L }\right) = y \alpha \triangle T $

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A uniform rod of length L and density $ \rho $ is being pulled along A smooth floor with a horizontal acceleration $ \alpha $ what is the magnitude of the stress at the transverse cross-section through the mid- point of the rod ?

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Explanation

$ F = M \alpha = ( L A \rho ) \alpha $ $ F- Fm = \left( { LA \rho \over 2 } \right) \alpha \Rightarrow F_m = \left( { LA \rho \over 2 } \right) \alpha $ stress at mid - point = $ { Fm \over A} = \left( { L\rho \alpha \over 2 } \right ) $

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