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Preeti Dabral 2 years, 8 months ago
The dimensional formula of energy or heat = u = [ ML2T-2 ].
Assume that, M1, L1, T1 and M2, L2, T2 are the units of mass, length and time in given two systems.
{tex}\therefore{/tex} M1 = 1 kg, M2 = {tex}\alpha{/tex} kg
L1 = 1 m, L2 = {tex}\beta{/tex} m
T1 = 1 s, T2 = {tex}\gamma{/tex} s
n1 and n2 are the magnitudes of heat or energy of system one and system two respectively.
For a given physical quantity, the product of its magnitude and unit is always constant.
n1u1 = n2u2
or {tex}n _ { 2 } = n _ { 1 } \frac { u _ { 1 } } { u _ { 2 } } = 4.2 \times \frac { \left[ \mathrm { M } _ { 1 } \mathrm { L } _ { 1 } ^ { 2 } \mathrm { T } _ { 1 } ^ { - 2 } \right] } { \left[ \mathrm { M } _ { 2 } \mathrm { L } _ { 2 } ^ { 2 } \mathrm { T } _ { 2 } ^ { - 2 } \right] }{/tex}
{tex}= 4.2 \left[ \frac { \mathrm { M } _ { 1 } } { \mathrm { M } _ { 2 } } \right] \times \left[ \frac { \mathrm { L } _ { 1 } } { \mathrm { L } _ { 2 } } \right] ^ { 2 } \times \left[ \frac { \mathrm { T } _ { 1 } } { \mathrm { T } _ { 2 } } \right] ^ { - 2 }{/tex}
{tex}= 4.2 \left[ \frac { 1 } { \alpha } \right] \times \left[ \frac { 1 } { \beta } \right] ^ { 2 } \times \left[ \frac { 1 } { \gamma } \right] ^ { - 2 }{/tex}
n2 = 4.2{tex}\alpha ^ { - 1 } \beta ^ { - 2 } \gamma ^ { 2 }{/tex} new unit
Therefore, 1 cal = {tex}4.2 \alpha ^ { - 1 } \beta ^ { - 2 } \gamma ^ { 2 }{/tex} new unit
Posted by Jahnvi Somani 2 years, 8 months ago
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Preeti Dabral 2 years, 8 months ago
Plan to go to a place by a vehicle. Take readings of odometer and speedometer after every 5 Minutes till you reach your destination. Record these observation in tabular form; plot graphs
Time (in Mins) Odometer (km) Speedometer (Km/Hr)
0 0 0
5 2 30
10 5 50
15 9 50
20 13 60
25 18 60
30 22 50
35 25 30
40 27 0
Motion is not uniform
Posted by Niharika Verma 3 years, 3 months ago
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Posted by Vamshik Shetty 2 years, 8 months ago
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Preeti Dabral 2 years, 8 months ago
{tex}\mathop {at\,0^\circ C}\limits^{ice} \to \mathop {at\,0^\circ C}\limits^{water} \to \mathop {at\,6^\circ C}\limits^{water} {/tex}
At equilibrium, heat released by 100gm of water = heat absorbed by 100gm of ice i.e.m1c1 (80 - 6) = m2 L + m2c1(6 - 0)[c1, m1, m2 and L are specific heat of water, mass of water, mass of ice and latent heat of fusion of ice respectively.)
{tex}\therefore{/tex}100 × 1 × 74 = 100 L + 100 × 1 × 6
{tex}\Rightarrow{/tex} L = (1 × 74) -6
{tex}\Rightarrow{/tex}L = 68 cal/gm
This is the required value of latent heat of fusion of ice.
Posted by Aanchal Sharma 2 years, 8 months ago
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Preeti Dabral 2 years, 8 months ago
In the steady-state,
Rate of flow of heat through cube A
= Rate of flow of heat through cube B.
or {tex}\frac{K_{1} A(100-T)}{x}=\frac{K_{2} A(T-0)}{x}{/tex}
or {tex}\frac{300 A(100-T)}{x}=\frac{200 A(T-0)}{x}{/tex}
or 300 - 3T = 2T or 5T = 300
T = 60°C
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Preeti Dabral 2 years, 8 months ago
I-shaped cross-section provides a large load bearing surface and enough depth to prevent buckling. Also it reduces the weight of the girder without sacrificing its strength.
Posted by Real Pradeep 2 years, 8 months ago
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Preeti Dabral 2 years, 8 months ago
The size of the mitochondria varies and the structure also varies according to the type of the cell. - Mitochondria is not like other cell organelles, where it is only visible if it is stained specifically.
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Madhu Dwivedi 3 years, 2 months ago
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