Yakeen NEET 3.0 Physics

<p>[A2V2D2] (A)</p> Signup and view all the answers

<p>[AV2D] (A)</p> Signup and view all the answers

<p>[A2VD] (D)</p> Signup and view all the answers

<p>ML–1T–1 (D)</p> Signup and view all the answers

<p>ML–1T–2 (C)</p> Signup and view all the answers

<p>FV–1 (C)</p> Signup and view all the answers

<p>T ∝ P^1D^2S^1 (C)</p> Signup and view all the answers

<p>1/2, –3/2, –1/2 (C)</p> Signup and view all the answers

<p>log(A) (C)</p> Signup and view all the answers

<p>ML–1T^0 (D)</p> Signup and view all the answers

<p>ML^0T^1 (C)</p> Signup and view all the answers

<p>M0L1T0 (A)</p> Signup and view all the answers

<p>ML–1T–2 (A)</p> Signup and view all the answers

<p>m/hr2 (D)</p> Signup and view all the answers

<p>y = 2A sin(2πx/λ) cos(2πct/λ) (C)</p> Signup and view all the answers

<p>The equation represents a progressive wave (C)</p> Signup and view all the answers

<p>m/hr (D)</p> Signup and view all the answers

<p>M0L1T–1 (B)</p> Signup and view all the answers

<p>Potential head (D)</p> Signup and view all the answers

The unit of 'a' is the inverse of time (T^-1). This is because 'at' must have the same dimensions as '3', which is dimensionless. Since 't' is in seconds, 'a' must be in units of per second or s^-1.

Work has the units of [ML^2T^-2]. If length, mass, and time are doubled, the unit of work increases by 8 times: [2M * (2L)^2 * (2T)^-2] = 8[ML^2T^-2].

Torque has dimensions of [ML^2T^-2]. It is the product of force and the perpendicular distance from the axis of rotation. This is equivalent to the dimensions of work or energy.

The dimensional representation of torque becomes [A^2V^2D^2] if the unit of length, mass, and time are each doubled. This is because torque is proportional to length squared and inversely proportional to the square of time.

The dimensional representation of torque becomes [AV^2D] if the unit of length is doubled. This is because torque is proportional to length.

The dimensional representation of torque becomes [A^2VD] if the unit of mass is doubled. This is because torque is proportional to mass.

The unit of x is [ML^-1T^-1]. This is because x = (ML^2T^-2) * T, which simplifies to [ML^-1T^-1].

The unit of 2πc/λ is [ML^-1T^-2]. This is because c is the speed of light with dimensions [LT^-1] and λ is wavelength with dimensions [L] 2π is dimensionless. So, the unit of 2πc/λ becomes [LT^-1]/[L] = [T^-1]. Multiplying by [ML^-1T^-1] gives [ML^-1T^-2].

The dimensions of mass (M) can be expressed in terms of energy (E), velocity (V), and force (F) as [FV^-1]. This is because Energy is equal to Force times Distance and Velocity is Distance divided by Time. Substituting the dimensions of Force, Distance, and Time, we derive the relationship: M = FV^-1.

The time period (T) is proportional to the product of pressure (P) raised to the power of 1/2, density (D) raised to the power of -3/2, and surface tension (S) raised to the power of -1/2. This is represented as T ∝ P^(1/2)D^(-3/2)S^(-1/2).

In the expression T ∝ P^aD^bS^c, a = 1/2, b = -3/2, and c = -1/2. This is found by analyzing the dimensions of each quantity and ensuring that dimensional consistency is maintained between the left and right sides of the equation.

The only dimensionally correct operation between two physical quantities A and B with different dimensions is taking the logarithm of A: log(A). This is because the logarithm of a quantity is dimensionless, regardless of the dimensions of the quantity itself.

The dimension of density (D) is [ML^-1T^0]. This is because density is defined as mass (M) per unit volume (L^3). So, the dimension of density is [M] / [L^3] = [ML^-3] = [ML^-1T^0].

The dimension of velocity (V) is [ML^0T^-1]. This is because velocity is defined as the rate of change of displacement (L) with respect to time (T). So, the dimension of velocity is [L] / [T] = [LT^-1] = [ML^0T^-1].

The constant on the right-hand side of Bernoulli's theorem is dimensionless. This is because it represents a potential head, which is a height and therefore has dimensions of length (L). Bernoulli's theorem states that the total energy per unit volume of a fluid is constant along a streamline, and this constant is expressed in units of length.

The dimensions of pressure gradient (dp/dx) are [ML^-1T^-2]. This is because pressure (p) has dimensions of [ML^-1T^-2] and distance (x) has dimensions of [L]. Therefore, the dimension of pressure gradient is [ML^-1T^-2] / [L] = [ML^-2T^-2].

The unit of 'b' is [m/hr^2]. This can be determined by ensuring dimensional consistency on both sides of the equation. Since x has units of meters (m) and t has units of hours (hr), bt^2 must also have units of meters. Therefore, 'b' must have units of [m/hr^2] to make the dimensions consistent.

The stationary wave equation is y = 2A sin(2πx/λ) cos(2πct/λ). This equation describes the displacement (y) of a point on a string as a function of position (x) and time (t). It represents a wave that appears to be standing still, resulting from the superposition of two identical waves traveling in opposite directions.

The statement that the equation represents a progressive wave is wrong. The equation represents a stationary wave, also known as a standing wave. Unlike a progressive wave, a stationary wave does not propagate through space, but instead oscillates in place.

The unit of 'a' is [m/hr]. This ensures dimensional consistency on both sides of the equation. Since x has units of meters (m) and t has units of hours (hr), 'at' must also have units of meters. Therefore, 'a' must have units of [m/hr] to make the dimensions consistent.

The constant on the left-hand side of Bernoulli's theorem has dimensions of [M0L1T-1]. This constant represents the velocity head, which is a velocity and therefore has dimensions of [LT^-1].

The constant on the right-hand side of Bernoulli's theorem represents the potential head. This is the height to which a fluid would rise if allowed to flow freely under the influence of gravity.