Tag: measurement of properties

Questions Related to measurement of properties

The international standards of weight and measures are made of:

  1. gold silver alloys

  2. platinum - iridium alloys

  3. copper - gold alloys

  4. platinum - iron alloys


Correct Option: B
Explanation:
The addition of 10% iridium to the platinum greatly increases its hardness and so reduces wear.Hence the international standards of weight and measures are made up of platinum and iridium alloys.

On Celsius scale, $0$ Kelvin is equal to:

  1. $273^0C$

  2. $0^oC$

  3. $-273^oC$

  4. $-373^oC$


Correct Option: C
Explanation:

To convert from Kelvin scale to Celsius scale, subtract 273 from Celsius scale. 

$0K=$$0-273=-273^oC$

What is the physical state of water at $-20^{\circ}C$?

  1. Solid

  2. Liquid

  3. Gas

  4. None of the above


Correct Option: A
Explanation:

Water at $-20^{\circ}C$ is ice so it is in solid state. It freezes at $0^{\circ}C$.

Convert 44 K into Celsius scale.

  1. $-233^oC$

  2. $-229^oC$

  3. $-212^oC$

  4. $-239^oC$


Correct Option: B
Explanation:

To convert from Kelvin scale to Celsius scale, subtract 273 from Kelvin scale, so $44-273=-229^oC$.


Hence, option $B$ is correct.

The centigrade $(^0 C)$ scale can be converted into Kelvin scale by the formula:

  1. Temperature in Kelvin + 273

  2. Temperature in Kelvin - 273

  3. Temperature in Fahrenheit + 273

  4. Temperature in Fahrenheit - 273


Correct Option: B
Explanation:

The temperature T in Kelvin (K) is equal to the temperature T in degrees Celsius $\displaystyle \left( ^{ \circ  }{ C } \right) + 273$: 


$\displaystyle { T } _{ \left( k \right)  }={ T } _{ \left( ^{ \circ  }{ C } \right)  }+273.15$

Therefore, temperature in $\displaystyle ^{ \circ  }{ C }$ =  temperature in Kelvin +273.

Hence, option B is correct.

Convert $30^oC$ to Kelvin scale.

  1. 323K

  2. 313 K

  3. 303 K

  4. 293 K


Correct Option: C
Explanation:

To convert from Celsius scale to Kelvin scale, we add 273 to Celsius scale. So, $30 + 273 = 303\ K$ 

Convert $-679^{\circ}$ F to celsius.

  1. 412.24

  2. $-$395

  3. 963

  4. $-$400


Correct Option: B
Explanation:

We know,    $\boxed { {  }^{ 0 }C=\left( {  }^{ 0 }F-32 \right) \times \frac { 5 }{ 9 }  } $

                   $\swarrow $               $\searrow $
temp in ${  }^{ 0 }C$                       temp in $F$.
${  }^{ 0 }C=\left( -679-32 \right) \times \dfrac { 5 }{ 9 } $
${  }^{ 0 }C=-395$
$\therefore $  temperature in ${  }^{ 0 }C$  u $-{ 395 }^{ 0 }C$.

What is the temperature of water in kelvin if the water is at $44^{\circ}C$ ?

  1. 330

  2. 317

  3. 229

  4. 230


Correct Option: B
Explanation:

Temperature in$ K$ = Temperature in $^{\circ}C + 273$


                                     = $44^{\circ}C + 273$

                                    = $317 K$.

So, the correct option is $B$

In the metric system, the prefix kilo-means :

  1. $10^0$

  2. $10^{-1}$

  3. $10^{-2}$

  4. $10^2$

  5. $10^3$


Correct Option: E
Explanation:

In the metric system, the prefix kilo-means $\displaystyle  10^3$. 
For example, 1 kg is equal to $\displaystyle  10^3 g$. 
1 km is equal to $\displaystyle  10^3 $ m.

Temperature of hot red iron is $58^{\circ}$ F. If the temperature increases by $20^{\circ}$ F, what is its temperature now in kelvin?

  1. 298.5

  2. 300

  3. 273

  4. $-$80


Correct Option: A
Explanation:

Temperature after heating $= 58 + 20 = 78^0F$
Temperature in Kelvin $= (F -  32) \div 1.8 + 273$
                                    $= (78-32) \div 1.8 +273$
                                    $= 46 \div 1.8 + 273$
                                    $= 25.55 + 273$
                                    $= 298.5$ K