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In an unsual vernier, 10 vernier scale divisions, coinside with 8 main scale divisions, thenwhat isthe least count of the vernier ?
$ 8M = 10V \Rightarrow 10M -2M = 10V , 10(M-V) = 2M (M -V ) = 2M /10 = 0.2 mm$
When the moment of force is maximum, then what is the angle between force and position vector of the force ?
$ \tau = Fr sin \theta $ $ ( sin \theta)_ { max} = 1 for \theta = 90 $
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 ?
$| \vec \tau | = | \vec r \times \vec F | $
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 ?
$ m _{max} = ( 2 kg ) ( {y_{max} \over x_{min}} ) = 12kg $
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.)
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
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 ?
Balancing moment $ 1x =2 (75 - x) \Rightarrow 3x = 150 or x = 50 cm$ Fulcrum is at 10 cm + 50 cm = 60 cm mark.
A liquid takes 5 minute to cool from $ 80 ^\circ C to 50 ^\circ C$ . The temperature of the surrounding is $ 20 ^\circ C$ . What is the time it will take to cool from $ 60 ^\circ C to 30 ^\circ C$ ?
Using the equation $ { \theta_1 - \theta _2 \over t } = K \left( {\theta_1 + \theta_2 \over 2} - \theta_0 \right) \theta_0$ = where = temperature of surrounding
Two spheres of the same material have radii 1 m and 4 m and temperatures 2000 k and 4000 k respectively. If the energy radiated by the spheres are $E_1 $ and $E_2 $ respectively then find ratio of $ { E_1 \over E_2 } $
$ { E_1 \over E_2 } = \left( { R_1 \over R_2 } \right)^2 \left( { T_1 \over T_4 } \right)^4$
A body cools in 5 minute from $ 60 ^\circ C to 40 ^\circ C $ .The temperature of the surroundings is $ 10 ^\circ C$ . What is its temperature after the next 5 minute ?
According to newton's law approximately, $ { \theta_1 - \theta _2 \over t } = K \left( {\theta_1 + \theta_2 \over 2} - \theta_0 \right) $ $ 160 - 4 \theta = 20 + \theta \Rightarrow \theta = 28 C $
What is the units of emissive power in stefan's law ?
Emissive power is defined as the radiant energy emitted per sec per unit area of the surface. Hence $ [E} = [ P/A] \Rightarrow unit of E = Wm^{-2 } $
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