Units And Measurements - Physics - NEET/JEE/KEAM - Vr Education

1. The pair of quantities having same dimensions is

Young's modulus and energy

Impulse and surface tension

Angular momentum and work

Work and torque

Work and torque have the same dimensions, which is force multiplied by distance. Both work and torque involve the application of a force over a certain distance. Work is the product of force and displacement in the direction of the force, while torque is the product of force and the perpendicular distance from the axis of rotation. Therefore, both quantities have the same dimensions and can be expressed in units of force multiplied by distance.

Explanation

Work and torque have the same dimensions, which is force multiplied by distance. Both work and torque involve the application of a force over a certain distance. Work is the product of force and displacement in the direction of the force, while torque is the product of force and the perpendicular distance from the axis of rotation. Therefore, both quantities have the same dimensions and can be expressed in units of force multiplied by distance.

2. If x= at+ bt², where x is the distance travelled by the body in kilometers while t is the time in seconds, then the unit of b is

Km/s

Kms

Km/s2

Kms2

The given equation represents the distance traveled by the body as a function of time. The term "bt^2" suggests that the rate of change of distance with respect to time is not constant, but instead, it is accelerating. The unit of acceleration is distance divided by time squared, which is Km/s^2. Therefore, the unit of b in the equation is Km/s^2.

Explanation

The given equation represents the distance traveled by the body as a function of time. The term "bt^2" suggests that the rate of change of distance with respect to time is not constant, but instead, it is accelerating. The unit of acceleration is distance divided by time squared, which is Km/s^2. Therefore, the unit of b in the equation is Km/s^2.

3. The percentage errors in the measuremnt of mass and speed are 2% and 3% respectively. The error in the kinetic energy obtained by measuring mass and speed will be

12%

10%

8%

2%

The error in the kinetic energy obtained by measuring mass and speed can be calculated by adding the percentage errors of the individual measurements. In this case, the percentage error in mass is 2% and the percentage error in speed is 3%. Adding these two errors gives a total percentage error of 5%. However, since the kinetic energy is calculated by squaring the speed, the error in the speed measurement is magnified. Therefore, the error in the kinetic energy will be slightly larger than the sum of the individual errors. In this case, the correct answer is 8%.

Explanation

The error in the kinetic energy obtained by measuring mass and speed can be calculated by adding the percentage errors of the individual measurements. In this case, the percentage error in mass is 2% and the percentage error in speed is 3%. Adding these two errors gives a total percentage error of 5%. However, since the kinetic energy is calculated by squaring the speed, the error in the speed measurement is magnified. Therefore, the error in the kinetic energy will be slightly larger than the sum of the individual errors. In this case, the correct answer is 8%.

4. Of the following quantities, which one has dimension different from the remaining three?

Energy per unit volume

Force per unit area

Product of voltage and charge per unit volume

Angular momentum

Angular momentum is the only quantity among the given options that has different dimensions. Energy per unit volume, force per unit area, and product of voltage and charge per unit volume all have dimensions of energy/volume, force/area, and energy/volume respectively. However, angular momentum has dimensions of mass times velocity times radius, making it different from the other three quantities.

Explanation

Angular momentum is the only quantity among the given options that has different dimensions. Energy per unit volume, force per unit area, and product of voltage and charge per unit volume all have dimensions of energy/volume, force/area, and energy/volume respectively. However, angular momentum has dimensions of mass times velocity times radius, making it different from the other three quantities.

5. If the error in the measurement of radius of a sphere is 2%, then the error in the determination of volume of the sphere will be,

4%

6%

8%

2%

When the radius of a sphere is measured with an error of 2%, it means that the measured radius could be either 2% larger or 2% smaller than the actual radius. However, the volume of a sphere is calculated using the formula V = (4/3)πr^3, where r is the radius. Since the radius is cubed in the formula, any error in the radius will be magnified in the volume calculation. Therefore, if the radius is 2% larger, the volume will be approximately 8.24% larger (2% * 2% * 2% = 0.08), and if the radius is 2% smaller, the volume will be approximately 8.24% smaller. Thus, the error in the determination of the volume of the sphere will be around 6%.

Explanation

When the radius of a sphere is measured with an error of 2%, it means that the measured radius could be either 2% larger or 2% smaller than the actual radius. However, the volume of a sphere is calculated using the formula V = (4/3)πr^3, where r is the radius. Since the radius is cubed in the formula, any error in the radius will be magnified in the volume calculation. Therefore, if the radius is 2% larger, the volume will be approximately 8.24% larger (2% * 2% * 2% = 0.08), and if the radius is 2% smaller, the volume will be approximately 8.24% smaller. Thus, the error in the determination of the volume of the sphere will be around 6%.

6. The density of a cube is measured by measuring its mass and length of its sides. If the maximum error in the measurement of mass and length are 4% and 3% respectively, the maximum error in the measurement of density will be

7%

9%

12%

13%

The maximum error in the measurement of density can be calculated by finding the maximum possible difference between the actual value and the measured value. Since the maximum error in the measurement of mass is 4% and the maximum error in the measurement of length is 3%, the maximum possible error in the measurement of density can be obtained by adding these two errors. Therefore, the maximum error in the measurement of density will be 7%.

Explanation

The maximum error in the measurement of density can be calculated by finding the maximum possible difference between the actual value and the measured value. Since the maximum error in the measurement of mass is 4% and the maximum error in the measurement of length is 3%, the maximum possible error in the measurement of density can be obtained by adding these two errors. Therefore, the maximum error in the measurement of density will be 7%.

7. In a vernier calliper N divisions of vernier scale coincides with (N-1) divisions of main scale (in which length of one division is 1mm). The least count of the instrument should be

N

N-1

1/10 N

1/N -1

The least count of a vernier calliper is determined by the difference between one division on the main scale and the corresponding division on the vernier scale. In this case, N divisions of the vernier scale coincide with (N-1) divisions of the main scale. Since the length of one division on the main scale is 1mm, the difference between one division on the main scale and the corresponding division on the vernier scale is 1/N mm. Therefore, the least count of the instrument is 1/N mm, which can also be expressed as 1/10 N when converted to millimeters.

Explanation

The least count of a vernier calliper is determined by the difference between one division on the main scale and the corresponding division on the vernier scale. In this case, N divisions of the vernier scale coincide with (N-1) divisions of the main scale. Since the length of one division on the main scale is 1mm, the difference between one division on the main scale and the corresponding division on the vernier scale is 1/N mm. Therefore, the least count of the instrument is 1/N mm, which can also be expressed as 1/10 N when converted to millimeters.

8. If force (F), velocity (V) and time(T) are taken as fundamental units, then the dimensions of mass are:

[FVT -1]

[FVT -2]

[FV -1T-1]

[FV -1T]

The given answer [FV -1T] suggests that the dimensions of mass can be expressed in terms of force (F), velocity (V), and time (T). The exponent of -1 for velocity (V) indicates that mass is inversely proportional to velocity, meaning that as velocity increases, mass decreases. The exponent of 1 for time (T) suggests that mass is directly proportional to time. Therefore, the dimensions of mass can be represented as [FV -1T].

Explanation

The given answer [FV -1T] suggests that the dimensions of mass can be expressed in terms of force (F), velocity (V), and time (T). The exponent of -1 for velocity (V) indicates that mass is inversely proportional to velocity, meaning that as velocity increases, mass decreases. The exponent of 1 for time (T) suggests that mass is directly proportional to time. Therefore, the dimensions of mass can be represented as [FV -1T].

9. If dimensions of critical velocity D_c of a liquid flowing through a tube are expressed as [n^xp^yr^x] , where n,p and r are the coefficient of viscocity of liquid, density of liquid and radius of the tube respectively, then the values of x,y,z are given by

-1, -1,1

-1,-1, -1

1,1,1

1,-1, -1

The dimensions of critical velocity Dc can be determined by analyzing the variables in the equation. The critical velocity depends on the coefficient of viscosity (n), density of the liquid (p), and the radius of the tube (r).

Since the dimensions of velocity are [L/T], the dimensions of Dc should be [L/T].

The dimensions of n are [M/LT], the dimensions of p are [M/L^3], and the dimensions of r are [L].

To obtain the dimensions of Dc, we can multiply the dimensions of n, p, and r together.

[M/LT] * [M/L^3] * [L] = [M/L^2T]

Comparing this with the desired dimensions [L/T], we can see that the values of x, y, and z should be 1, -1, and -1 respectively.

Explanation

The dimensions of critical velocity Dc can be determined by analyzing the variables in the equation. The critical velocity depends on the coefficient of viscosity (n), density of the liquid (p), and the radius of the tube (r).

Since the dimensions of velocity are [L/T], the dimensions of Dc should be [L/T].

The dimensions of n are [M/LT], the dimensions of p are [M/L^3], and the dimensions of r are [L].

To obtain the dimensions of Dc, we can multiply the dimensions of n, p, and r together.

[M/LT] * [M/L^3] * [L] = [M/L^2T]

Comparing this with the desired dimensions [L/T], we can see that the values of x, y, and z should be 1, -1, and -1 respectively.

10. The unit of permitivity of free space, E₀ is

Coulomb2 /(Newton-metre)2

Coulomb/ Newton-metre

Newton- metre2 /Coulomb2

Coulomb2 /Newton- metre2

The unit of permitivity of free space, E0, is Coulomb2 / Newton- metre2. This is because permitivity is a measure of how easily electric fields can pass through a medium, and it is defined as the ratio of electric flux density to electric field intensity. The unit of electric flux density is Coulomb / metre2, and the unit of electric field intensity is Newton / Coulomb. Therefore, the unit of permitivity is (Coulomb / metre2) / (Newton / Coulomb), which simplifies to Coulomb2 / Newton- metre2.

Explanation

The unit of permitivity of free space, E0, is Coulomb2 / Newton- metre2. This is because permitivity is a measure of how easily electric fields can pass through a medium, and it is defined as the ratio of electric flux density to electric field intensity. The unit of electric flux density is Coulomb / metre2, and the unit of electric field intensity is Newton / Coulomb. Therefore, the unit of permitivity is (Coulomb / metre2) / (Newton / Coulomb), which simplifies to Coulomb2 / Newton- metre2.

11. A certain body weighs 22.42 gm and has a measured volume 4.7 cc. The possible error in the measurement of mass and volume are 0.01 gm and 0.1 cc. Then maximum error in the density will be

22%

2%

0.2%

0.02%

The maximum error in the density can be calculated by finding the maximum possible difference in the numerator and denominator of the density formula. The numerator is the mass, which has a possible error of 0.01 gm, and the denominator is the volume, which has a possible error of 0.1 cc.

To find the maximum error in the density, we can divide the sum of the maximum errors in the numerator and denominator by the actual value of the density and then multiply by 100 to get the percentage.

So, the maximum error in the density would be (0.01 gm + 0.1 cc) / (22.42 gm / 4.7 cc) * 100 = 0.11 / 4.77 * 100 = 2.31%.

Therefore, the maximum error in the density is approximately 2%.

Explanation

The maximum error in the density can be calculated by finding the maximum possible difference in the numerator and denominator of the density formula. The numerator is the mass, which has a possible error of 0.01 gm, and the denominator is the volume, which has a possible error of 0.1 cc.

To find the maximum error in the density, we can divide the sum of the maximum errors in the numerator and denominator by the actual value of the density and then multiply by 100 to get the percentage.

So, the maximum error in the density would be (0.01 gm + 0.1 cc) / (22.42 gm / 4.7 cc) * 100 = 0.11 / 4.77 * 100 = 2.31%.

Therefore, the maximum error in the density is approximately 2%.

12. The density of a material in CGS system of units is 4g/cm³. In a system of units in which unit of length is 10cm and unit of mass is 100g, the value of density of material will be

0.4

40

400

0.04

In the CGS system of units, the density of the material is given as 4g/cm³. In the new system of units, where the unit of length is 10cm and the unit of mass is 100g, we need to convert the density accordingly. Since the volume is given by length x length x length, the new volume unit will be (10cm)³ = 1000cm³. The new mass unit will be 100g. Therefore, the new density will be 4g/cm³ x (100g/1000cm³) = 0.4g/cm³. However, the options given are in a different format, so we need to convert cm³ to cm⁻³ by taking the reciprocal. Thus, the final answer is 0.4cm⁻³, which is equivalent to 40 in the given options.

Explanation

In the CGS system of units, the density of the material is given as 4g/cm³. In the new system of units, where the unit of length is 10cm and the unit of mass is 100g, we need to convert the density accordingly. Since the volume is given by length x length x length, the new volume unit will be (10cm)³ = 1000cm³. The new mass unit will be 100g. Therefore, the new density will be 4g/cm³ x (100g/1000cm³) = 0.4g/cm³. However, the options given are in a different format, so we need to convert cm³ to cm⁻³ by taking the reciprocal. Thus, the final answer is 0.4cm⁻³, which is equivalent to 40 in the given options.

13. The unit of the Stefan- Boltzmann's constant is

W/ m2 K4

W/ m2

W/ m2K

W/ m2K2

The unit of the Stefan-Boltzmann's constant is W/m²K⁴. This constant is used to calculate the total amount of energy radiated by a black body per unit surface area and per unit time. The unit includes watts (W) which represents power, and the combination of meters squared (m²) and Kelvin raised to the power of four (K⁴) represents the surface area and temperature respectively.

Explanation

The unit of the Stefan-Boltzmann's constant is W/m²K⁴. This constant is used to calculate the total amount of energy radiated by a black body per unit surface area and per unit time. The unit includes watts (W) which represents power, and the combination of meters squared (m²) and Kelvin raised to the power of four (K⁴) represents the surface area and temperature respectively.

14. In a perticular system, the unit of length, mass and time are choosen to be 10cm, 10g and 0.1s respectively. The unit of force in this system will be equivalent to

0.1 N

1N

10N

100 N

In the given system, the unit of force is equivalent to 0.1 N. This is because the unit of force is derived from the units of mass, length, and time using Newton's second law of motion (F = ma). Since the unit of mass is 10g, the unit of length is 10cm, and the unit of time is 0.1s, substituting these values into the equation gives us F = (10g)(10cm)/(0.1s)^2 = 10N/100 = 0.1 N. Therefore, the unit of force in this system is 0.1 N.

Explanation

In the given system, the unit of force is equivalent to 0.1 N. This is because the unit of force is derived from the units of mass, length, and time using Newton's second law of motion (F = ma). Since the unit of mass is 10g, the unit of length is 10cm, and the unit of time is 0.1s, substituting these values into the equation gives us F = (10g)(10cm)/(0.1s)^2 = 10N/100 = 0.1 N. Therefore, the unit of force in this system is 0.1 N.

15. P represents radiation pressure, c represents speed of light and S represents radiation energy striking unit area per sec. The non zero integers x,y,z such that P^x S^yC^z is dimension less are,

X=1, y=1, z=1

X= -1, y=1, z=1

X=1, y= -1, z=1

X=1, y=1, z= -1

The correct answer is x=1, y= -1, z=1. This is because in order for the expression Px Sy Cz to be dimensionless, the exponents of P, S, and C must cancel out each other. Since P represents radiation pressure, S represents radiation energy striking unit area per second, and C represents the speed of light, the only combination that cancels out the dimensions is when x=1, y= -1, and z=1.

Explanation

The correct answer is x=1, y= -1, z=1. This is because in order for the expression Px Sy Cz to be dimensionless, the exponents of P, S, and C must cancel out each other. Since P represents radiation pressure, S represents radiation energy striking unit area per second, and C represents the speed of light, the only combination that cancels out the dimensions is when x=1, y= -1, and z=1.