Class 10 Scientific Learning Exercise 1.1 Solutions | Science and Technology Curriculum Development Centre

Option C is the correct answer.

Newton, pascal, and joule are the units of force, pressure, and energy, which are derived physical quantities that depend upon fundamental physical quantities.

Kilogram is a unit of mass, which is a fundamental physical quantity.

Option B is the correct answer.

Velocity has the unit $\rm ms^{-1}$.

We know, $\rm velocity = \frac{displacement}{time taken}$

The fundamental unit of displacement is ‘m’ and that of time is ‘s’.

Thus, the unit of velocity is $\rm \frac{m}{s} = ms^{-1}$.

On the other hand, the units of acceleration, force, and density are $\rm ms^{-2}$, $\rm kgms^{-2}$, and $\rm kgm^{-3}$, respectively.

Option C is the correct answer.

Candela, ampere, and kelvin are the units of Luminous Intensity, Current, and Temperature. These three quantities are fundamental physical quantities. They do not depend upon any other physical quantity.

On the other hand, Joule is the unit of Work done or Energy, which is a derived physical quantity. It depends upon mass (kg), length (m), and time (s).

Hence, Joule is the derived unit among the given four units.

Option A is the correct answer.

Newton is the unit of force.

By a generalization of Newton's second law, we know,

Force (F) = mass (m) x acceleration (a)

The unit of mass is kilogram $\rm (kg)$ and acceleration is $\rm (ms^{-2})$.

So, F = $\rm kg \times ms^{-2}$

$\rm = kgms^{-2}$

Work is a derived physical quantity. Its unit is Joule.

Mathematically, Work = Force x displacement.

So, Joule is the product of Newton (N) and metre (m). Again, Newton is a derived unit of Force and is equal $\rm kgms^{-2}$.

Hence,

Joule (J) = $\rm kgms^{-2} \cdot m = kg m^2s^{-2}$.

Clearly, Joule depends upon three fundamental physical quantities, i.e., kilogram (kg), metre (m), and second (s). Hence, it is the derived unit of work.

More than two variables (dependent and independent) can influence the results of an experiment. To increase an experiment's accuracy and reliability and eliminate any form of inconsistencies in the experiment, controlled variables are important.

To test the validity of a physical equation, we perform unit-wise analysis. In the above physical relation:

$$\rm v^2 = ut$$

In terms of fundamental physical units,

$$\rm (ms^{-2})^2 \neq ms^{-1} \cdot s$$

The fundamental physical units involved in the equation's left-hand sides are not equal to those in the right-hand side of the equation. So, we can conclude that the given relation $\rm v^2 = ut$ is invalid.

The differences between Fundamental units and Derived units are mentioned in the table below:

A unit is a standard physical quantity that is taken as a reference to measure other physical quantities of the same kind.

Example: kilogram is the unit of mass. A box is two kilograms means that the box is twice as heavy as the standard one-kilogram unit.

Following are the SI units of the given quantities:

• mass: kilogram (kg)
• temperature: kelvin (K)
• energy: joule (J)
• density: kilogram per cubic-metre (kg/m$^3$).

We can check the validity of an equation by using unit analysis of the physical quantities involved in the equation. In the analysis of the unit-wise equation, we check the units of each term involved in the equation. If each term has the same units, the given equation is validated as true, otherwise false.

For example, an equation $\rm v = u + at$ is a valid physical equation. This is because the units of $\rm v$ is $\rm ms^{-1}$, $\rm u$ is $\rm ms^{-1}$, and $\rm at$ is $\rm ms^{-2} \cdot s = ms^{-1}$. Each term has the same units. Hence, the given equation is valid.

Kilogram (kg), metre (m), and second (s) are the fundamental units involved in the units of pressure.

We know,

$$\rm Pressure = \frac{Force}{Area}$$

$$\rm P = \frac{ kgms^{-2} }{ m^2 }$$

$$\rm P = kgm^{-1}s^{-2}$$

To find the fundamental units involved in the given derived units, we first write their corresponding fundamental physical quantities and definitions.

i) newton (N)

Newton is the unit of Force. And, $$\rm Force = mass \times acceleration$$

$$\rm N = kg \cdot ms^{-2}$$

Therefore, newton involves three fundamental units: kilogram (kg), meter (m), and second (s).

ii) watt (W)

Watt is the unit of Power. And, $$\rm Power = Force \times velocity$$

$$\rm W = kgms^{-2} \cdot ms^{-1}$$

Therefore, watt involves three fundamental units: kilogram (kg), meter (m), and second (s).

iii) joule (J)

Joule is the unit of Work. And, $$\rm Work = Force \times distance$$

$$\rm J = kgms^{-2} \cdot m$$

Therefore, joule involves three fundamental units: kilogram (kg), meter (m), and second (s).

iv) pascal (Pa)

Pascal is the unit of Pressure. And $$\rm Pressure = \frac{Force}{Area}$$

$$\rm Pa = \frac{kgms^{-2}}{m^2}$$

Therefore, pascal involves three fundamental units: kilogram (kg), meter (m), and second (s).

Checking the validity of formula $\rm P = mv^{2}$.

STEP 1: Find the units of the physical quantities on the left-hand side of the equation.

$\rm P = \frac{Work}{Time} = \frac{kgm^{2}s^{-2}}{s} = kgm^{2}s^{-3}$

STEP 2: Find the units of the physical quantities on the right-hand side of the equation,

$\rm mv^{2} = kg \cdot \left ( \frac{m}{s} \right ) ^{2} = kgm^{2} s^{-2}$

From steps 1 and 2, we observe that the units of the terms in the given equation are different. Hence, this alternative formula for Power is invalid.

Checking the validity of formula $\rm P = \frac{mv}{A}$.

STEP 1: Find the units of the physical quantities on the left-hand side of the equation.

$\rm P = \frac{Force}{Area} = \frac{kgms^{-2}}{m^{2}} = kgm^{-1}s^{-2}$

STEP 2: Find the units of the physical quantities on the right-hand side of the equation,

$\rm \frac{mv}{A} = \frac{kg \cdot ms^{-1}}{m^{2}} = kg m^{-1} s^{-1}$

From steps 1 and 2, we observe that the units of the terms in the given equation are different. Hence, this alternative formula for Pressure is also invalid.

Independent Variable: This is the variable that is manipulated or changed in an experiment to observe its effect on the dependent variable.

• Example: In a study on plant growth, the amount of sunlight each plant receives (e.g., 2 hours, 4 hours, 6 hours) is the independent variable.

Dependent Variable: This variable is measured or observed in response to changes in the independent variable. It depends on the independent variable.

• Example: In the same plant growth study, the height of the plants after a certain period is the dependent variable, as it depends on the amount of sunlight they received.

Controlled Variable: These are the variables that are kept constant throughout the experiment to ensure that the results are due to the independent variable alone.

• Example: In the plant growth experiment, controlled variables could include the type of plant, the soil type, water amount, and temperature, all of which should remain the same for all plants to ensure a fair test.

In this experiment, Karma studies the effect of the thickness of wire (causative factor) on the life span of a dry cell (effect). Hence, we can list the variables of scientific research as follows:

• Independent variable: Thickness of the wire
• Dependent variable: Life span of dry cell
• Controlled variable: Power of bulb, Number of bulbs, Potential difference of the dry cell, Previous usage of the dry cell, and more.

Chandani's experiment has all three variables of scientific research. We have identified them and noted below:

• Independent Variable: Amount of substances mixed into the soil, including lime, urea fertilizer, common salt, and compost manure
• Dependent Variable: Height of plant
• Controlled Variable: Seeds of the plant, Quantity of Soil, Amount of sunlight, Water received by the plant

By using three pots for her experiment, Chandani minimizes error and eliminates anomalies. If it is true that the amount of substances mixed into the soil affects the growth (in height) of the plant, the seeds placed in the three pots containing the same amount of substances must experience similar growth.

This factor increases the validity of the outcomes of the experiment.

Dependent Variable: The expected outcome of the experiment is to find out how the color of an object affects its ability to hold heat. So, it is evident that the measured temperature of water in each flash is the dependent variable.

Independent Variable: The variation that is brought about in four of the cans is their colours. Hence, the colour of the conical flasks is the independent variable.

Controlled Variable:

• Sunlight: What affects the rise or fall in the temperature of water in the flask is the amount of heat received from the sun. Subodh can control the duration for which the conical flasks are kept in the sunlight. He can also adjust the direction of sunlight falling on the flasks.
• Water: The quantity and purity of water used in the experiment is also a controlled variable.
• Other factors such as the specific heat capacity of a substance, its area, and mass also affect the quantity of heat (to be studied later in Heat, Q = ms 'dt'). The material used to make the flask is a controlled variable. Furthermore, its dimensions are controlled.

The controlled variables must be the same for all of the four conical flasks. Otherwise, Subodh may observe errors in his experiment.

Remember, Subodh must also consider other factors. For instance, the flasks should be kept on insulating surfaces. The thickness of the colour coating must be uniform. The thermometer used to measure the temperature of the water in all four flasks must have the same precision and accuracy.

Learn more about Dependent, Independent, and Controlled Variables.

In Manisha's experiment to test the eating habits of her dog, we can identify the following variables of scientific research:

• Independent variable: amount of food, time of giving food
• Dependent variable: the speed at which the dog ate

An experiment with two independent variables cannot correctly and reliably study the effects of these variables on a third variable. Hence, to correct her experiment, Manisha should do either of the following:

• Independent variable: amount of food; Controlled variable: time of giving food
• Independent variable: time of giving food; Controlled variable: amount of food

By doing so, she can correctly study the influence of either of these factors on the speed at which the dog ate.

Ohm ($\rm \Omega$) is the unit of electric resistance. We define electric resistance as the ratio of Voltage to Current, i.e., $\rm R = \frac{V}{I}$.

We know that Voltage (or potential difference) is Work done and Charge, i.e., $\rm V = \frac{W}{q}$.

When we combine the above two equations, we get,

$\rm R = \frac{W}{q \cdot I}$

Expressing charge in the form of electric current, we get,

$\rm R = \frac{W}{ \frac{q}{T} \cdot {T} \cdot I}$

$\rm R = \frac{W}{ I \cdot T \cdot I}$

$\rm R = \frac{W}{ I^{2} \cdot T}$

Replacing the symbols with their corresponding units, we get,

$\rm \Omega = \frac{ kg m^{2} s^{-2} }{ A^{2} \cdot s}$

$\rm or, \Omega= kg m^{2} s^{-2 -1} A^{-2}$

$\rm \therefore \Omega = kg m^{2}s^{-3} A^{-2}$

Hence, the unit of electric resistance ($\Omega$) is $\rm kg m^{2}s^{-3} A^{-2}$.