Maths-

General

Easy

### Question

#### divided by 4 in fraction

#### The correct answer is:

#### Convert the mixed fraction into improper fractions.

= 8 × 3 + =

First, To divide fractions is to find the reciprocal (reverse the numerator and denominator) of the second fraction and then multiply the two numerators. Then, multiply the two denominators.

= x =

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### Related Questions to study

Maths-

#### Decimal equivalent of 2 1/8 is:

Convert the fraction into improper form then divide

2 1/8 = 8*2+1/8= 17/8 by dividing numerator by denominator it is equal to 2.125

2 1/8 = 8*2+1/8= 17/8 by dividing numerator by denominator it is equal to 2.125

#### Decimal equivalent of 2 1/8 is:

Maths-General

Convert the fraction into improper form then divide

2 1/8 = 8*2+1/8= 17/8 by dividing numerator by denominator it is equal to 2.125

2 1/8 = 8*2+1/8= 17/8 by dividing numerator by denominator it is equal to 2.125

Maths-

#### 5/6 can be represented in percentage as:

First divide the numerator by denominator

5/6= 0.833

And then convert fraction into percent we multiply it by 100 and divide it by 100.

0.833*100/100= 83.33/100

So 83.33 per cent that means 83.33%

5/6= 0.833

And then convert fraction into percent we multiply it by 100 and divide it by 100.

0.833*100/100= 83.33/100

So 83.33 per cent that means 83.33%

#### 5/6 can be represented in percentage as:

Maths-General

First divide the numerator by denominator

5/6= 0.833

And then convert fraction into percent we multiply it by 100 and divide it by 100.

0.833*100/100= 83.33/100

So 83.33 per cent that means 83.33%

5/6= 0.833

And then convert fraction into percent we multiply it by 100 and divide it by 100.

0.833*100/100= 83.33/100

So 83.33 per cent that means 83.33%

Physics-

#### Two holes of unequal diameters d_{1} and d_{2} are cut in a metal sheet. If the sheet is heated

If the sheet is heated then both d

_{1}and d_{2}will increase since the thermal expansion of isotropic solid is similar to true photographic enlargement#### Two holes of unequal diameters d_{1} and d_{2} are cut in a metal sheet. If the sheet is heated

Physics-General

If the sheet is heated then both d

_{1}and d_{2}will increase since the thermal expansion of isotropic solid is similar to true photographic enlargementPhysics-

#### Out of the following, in which vessel will the temperature of the solution be higher after the salt is completely dissolved.

When salt crystals dissolves, crystal lattice is destroyed. The process requires a certain amount of energy (latent heat) which is taken from the water.

In vessel

In vessel

#### Out of the following, in which vessel will the temperature of the solution be higher after the salt is completely dissolved.

Physics-General

When salt crystals dissolves, crystal lattice is destroyed. The process requires a certain amount of energy (latent heat) which is taken from the water.

In vessel

In vessel

Physics-

#### Two substances A and B of equal mass m are heated at uniform rate of 6 cal s^{–1} under similar conditions. A graph between temperature and time is shown in figure. Ratio of heat absorbed by them for complete fusion is

From given curve,

Melting point for

and melting point for

Time taken by A for fusion minute

Time taken by B for fusion minute

Then .

Melting point for

and melting point for

Time taken by A for fusion minute

Time taken by B for fusion minute

Then .

#### Two substances A and B of equal mass m are heated at uniform rate of 6 cal s^{–1} under similar conditions. A graph between temperature and time is shown in figure. Ratio of heat absorbed by them for complete fusion is

Physics-General

From given curve,

Melting point for

and melting point for

Time taken by A for fusion minute

Time taken by B for fusion minute

Then .

Melting point for

and melting point for

Time taken by A for fusion minute

Time taken by B for fusion minute

Then .

Physics-

#### Which of the substances A, B or C has the highest specific heat ? The temperature vs time graph is shown

Substances having more specific heat take longer time to get heated to a higher temperature and longer time to get cooled.

If we draw a line parallel to the time axis then it cuts the given graphs at three different points. Corresponding points on the times axis shows that

Þ

If we draw a line parallel to the time axis then it cuts the given graphs at three different points. Corresponding points on the times axis shows that

Þ

#### Which of the substances A, B or C has the highest specific heat ? The temperature vs time graph is shown

Physics-General

Substances having more specific heat take longer time to get heated to a higher temperature and longer time to get cooled.

If we draw a line parallel to the time axis then it cuts the given graphs at three different points. Corresponding points on the times axis shows that

Þ

If we draw a line parallel to the time axis then it cuts the given graphs at three different points. Corresponding points on the times axis shows that

Þ

Physics-

#### The graph signifies

#### The graph signifies

Physics-General

Physics-

#### A student takes 50gm wax (specific heat = 0.6 kcal/kg^{°}C) and heats it till it boils. The graph between temperature and time is as follows. Heat supplied to the wax per minute and boiling point are respectively

Since specific heat = 0.6 kcal/gm ´ °C = 0.6 cal/gm ´°C

From graph it is clear that in a minute, the temperature is raised from 0°C to 50°C.

Þ Heat required for a minute = 50 ´ 0.6 ´ 50 = 1500 cal.

Also from graph, Boiling point of wax is 200°C.

From graph it is clear that in a minute, the temperature is raised from 0°C to 50°C.

Þ Heat required for a minute = 50 ´ 0.6 ´ 50 = 1500 cal.

Also from graph, Boiling point of wax is 200°C.

#### A student takes 50gm wax (specific heat = 0.6 kcal/kg^{°}C) and heats it till it boils. The graph between temperature and time is as follows. Heat supplied to the wax per minute and boiling point are respectively

Physics-General

Since specific heat = 0.6 kcal/gm ´ °C = 0.6 cal/gm ´°C

From graph it is clear that in a minute, the temperature is raised from 0°C to 50°C.

Þ Heat required for a minute = 50 ´ 0.6 ´ 50 = 1500 cal.

Also from graph, Boiling point of wax is 200°C.

From graph it is clear that in a minute, the temperature is raised from 0°C to 50°C.

Þ Heat required for a minute = 50 ´ 0.6 ´ 50 = 1500 cal.

Also from graph, Boiling point of wax is 200°C.

Physics-

#### Heat is supplied to a certain homogenous sample of matter, at a uniform rate. Its temperature is plotted against time, as shown. Which of the following conclusions can be drawn

The horizontal parts of the curve, where the system absorbs heat at constant temperature must depict changes of state. Here the latent heats are proportional to lengths of the horizontal parts. In the sloping parts, specific heat capacity is inversely proportional to the slopes.

#### Heat is supplied to a certain homogenous sample of matter, at a uniform rate. Its temperature is plotted against time, as shown. Which of the following conclusions can be drawn

Physics-General

The horizontal parts of the curve, where the system absorbs heat at constant temperature must depict changes of state. Here the latent heats are proportional to lengths of the horizontal parts. In the sloping parts, specific heat capacity is inversely proportional to the slopes.

Physics-

#### Which of the curves in figure represents the relation between Celsius and Fahrenheit temperatures

Þ Hence graph between °C and °F will be a straight line with positive slope and negative intercept.

#### Which of the curves in figure represents the relation between Celsius and Fahrenheit temperatures

Physics-General

Þ Hence graph between °C and °F will be a straight line with positive slope and negative intercept.

Physics-

#### A solid substance is at 30°C. To this substance heat energy is supplied at a constant rate. Then temperature versus time graph is as shown in the figure. The substance is in liquid state for the portion (of the graph)

In the given graph CD represents liquid state.

#### A solid substance is at 30°C. To this substance heat energy is supplied at a constant rate. Then temperature versus time graph is as shown in the figure. The substance is in liquid state for the portion (of the graph)

Physics-General

In the given graph CD represents liquid state.

Physics-

#### The figure given below shows the cooling curve of pure wax material after heating. It cools from A to B and solidifies along BD. If L and C are respective values of latent heat and the specific heat of the liquid wax, the ratio L/C is

Let the quantity of heat supplied per minute be Q. Then quantity of heat supplied in 2 min

In 4 min, heat supplied

\ Þ

In 4 min, heat supplied

\ Þ

#### The figure given below shows the cooling curve of pure wax material after heating. It cools from A to B and solidifies along BD. If L and C are respective values of latent heat and the specific heat of the liquid wax, the ratio L/C is

Physics-General

Let the quantity of heat supplied per minute be Q. Then quantity of heat supplied in 2 min

In 4 min, heat supplied

\ Þ

In 4 min, heat supplied

\ Þ

Physics-

#### The portion AB of the indicator diagram representing the state of matter denotes

The volume of matter in portion AB of the curve is almost constant and pressure is decreasing. These are the characteristics of liquid state.

#### The portion AB of the indicator diagram representing the state of matter denotes

Physics-General

The volume of matter in portion AB of the curve is almost constant and pressure is decreasing. These are the characteristics of liquid state.

Physics-

#### The graph shows the variation of temperature (T) of one kilogram of a material with the heat (H) supplied to it. At O, the substance is in the solid state. From the graph, we can conclude that

Since in the region AB temperature is constant therefore at this temperature phase of the material changes from solid to liquid and (H

Similarly in the region CD temperature is constant therefore at this temperature phase of the material changes from liquid to gas and (H

_{2}– H_{1}) heat will be absorb by the material. This heat is known as the heat of melting of the solid.Similarly in the region CD temperature is constant therefore at this temperature phase of the material changes from liquid to gas and (H

_{4}– H_{3}) heat will be absorb by the material. This heat as known as the heat of vaporisation of the liquid.#### The graph shows the variation of temperature (T) of one kilogram of a material with the heat (H) supplied to it. At O, the substance is in the solid state. From the graph, we can conclude that

Physics-General

Since in the region AB temperature is constant therefore at this temperature phase of the material changes from solid to liquid and (H

Similarly in the region CD temperature is constant therefore at this temperature phase of the material changes from liquid to gas and (H

_{2}– H_{1}) heat will be absorb by the material. This heat is known as the heat of melting of the solid.Similarly in the region CD temperature is constant therefore at this temperature phase of the material changes from liquid to gas and (H

_{4}– H_{3}) heat will be absorb by the material. This heat as known as the heat of vaporisation of the liquid.Physics-

#### The graph AB shown in figure is a plot of temperature of a body in degree celsius and degree Fahrenheit. Then

Relation between Celsius and Fahrenheit scale of temperature is Þ C =

Equating above equation with standard equation of line we get slope of the line AB is

Equating above equation with standard equation of line we get slope of the line AB is

#### The graph AB shown in figure is a plot of temperature of a body in degree celsius and degree Fahrenheit. Then

Physics-General

Relation between Celsius and Fahrenheit scale of temperature is Þ C =

Equating above equation with standard equation of line we get slope of the line AB is

Equating above equation with standard equation of line we get slope of the line AB is