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A calorie is a unit of heat or energy and it equals about 4.2 J where 1J = 1 kg m2 sñ2. Suppose we employ a system of units in which the unit of mass equals αkg, the unit of length equals β m, the unit of time is γ s. Show that a calorie has a magnitude 4.2 straight alpha to the power of negative 1 end exponent straight beta to the power of negative 1 end exponent straight gamma squared in terms of the new units. 


We know that dimensional formula of heat energy is open square brackets straight M to the power of 1 straight L squared straight T to the power of negative 2 end exponent close square brackets

It is given that,

1 cal = 4.2J 

In SI system,   

M= 1 kg, L= 1 m,  T= 1 s 

In new system of units, 

straight M subscript 2 equals straight alpha space kg comma space straight L subscript 2 equals straight beta space straight m comma space straight T subscript 2 equals straight gamma space straight s

We  know,  

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#6 {main}</pre> 

space space space space straight n subscript 2 equals 4.2 open parentheses fraction numerator 1 space kg over denominator straight alpha space kg end fraction close parentheses to the power of 1 open parentheses fraction numerator 1 space straight m over denominator straight beta space straight m end fraction close parentheses squared open parentheses fraction numerator 1 space straight s over denominator straight gamma space straight s end fraction close parentheses to the power of negative 2 end exponent

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#6 {main}</pre>

therefore space space space space space space space space 1 space cal space equals space equals 4.2 straight alpha to the power of negative 1 end exponent straight beta to the power of negative 2 end exponent straight gamma squared space new space unit 

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Fill in the blanks : 
c) A vehicle moving with a speed of 18 km h–1covers....m in 1 s




Tips: -

1 space km divided by hr space equals space fraction numerator 1 cross times 1000 over denominator 3600 end fraction space equals space 5 over 18 space straight m divided by straight s space
straight i. straight e. comma space 18 space km divided by hr space equals space 18 cross times 5 over 18 equals space 5 space straight m divided by straight s space

Now comma space

Distance space equals space speed space cross times space time space

space space space space space space space space space space space space space space space space space equals space 5 cross times 1 space equals space 5 space straight m
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The surface area of a solid cylinder of radius 2.0 cm and height 10.0 cm is equal to
...(mm)2

1.5 x 104 mm2

The total surface area of a cylinder of radius r and height h is

S = 2πr (r + h).

Given space that comma space

straight r space equals space 2 space cm space equals space 2 cross times 10 space mm space equals space 20 space mm space

straight h space equals space 10 space cm space equals 10 space cross times 10 space mm space equals space 100 space mm space

So comma space

Surface space area comma space straight S space equals space 2 cross times 3.14 cross times 20 cross times left parenthesis 20 plus 100 right parenthesis

space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals space 15072 space

space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals space 1.5 space cross times 10 to the power of 4 space mm squared

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State the number of significant figures in the following : (a) 0.007 m (b) 2.64 cross times space 10 to the power of 24 kg (c) 0.2370 g cm-3 d) 6.320 J e)   6.032 N m-
f)  0.0006032 m2


The number of significant figures are: 

(a) One (b) Three (c) Four d) Three e) Four f) seven
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Fill in the blanks: 
a) The volume of a cube of side 1 cm is equal to .....m3


10-6 

1 cm = 1 space cm space equals space 1 over 100 space straight m
So comma space volume space of space the space cube space equals space 1 space cm cubed space

straight i. straight e. comma space 1 space cm cubed space equals space open parentheses 1 over 100 close parentheses cross times space open parentheses 1 over 100 close parentheses cross times space open parentheses 1 over 100 close parentheses space straight m space

therefore space 1 space cm cubed space equals space 10 to the power of negative 6 end exponent space straight m cubed

So space volume space of space straight a space cube space of space side space 1 space cm
is space equal space to space 10 to the power of negative 6 end exponent space cm cubed.

223 Views