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Question: If radius of the nucleus is 3.5 × 10⁻¹⁵ m then if the space or volume occupied by the nucleus is y ×...

If radius of the nucleus is 3.5 × 10⁻¹⁵ m then if the space or volume occupied by the nucleus is y × 10⁻⁴³ m³, the value of 10y is......... [Nearest Integer]

Answer

18

Explanation

Solution

The volume of a nucleus, which is approximated as a sphere, is given by the formula: V=43πR3V = \frac{4}{3}\pi R^3

Given the radius of the nucleus, R=3.5×10−15R = 3.5 \times 10^{-15} m.

First, calculate the cube of the radius: R3=(3.5×10−15 m)3R^3 = (3.5 \times 10^{-15} \text{ m})^3 R3=(3.5)3×(10−15)3 m3R^3 = (3.5)^3 \times (10^{-15})^3 \text{ m}^3 R3=42.875×10−45 m3R^3 = 42.875 \times 10^{-45} \text{ m}^3

Now, substitute this value into the volume formula: V=43π(42.875×10−45 m3)V = \frac{4}{3}\pi (42.875 \times 10^{-45} \text{ m}^3) V=4×42.8753π×10−45 m3V = \frac{4 \times 42.875}{3} \pi \times 10^{-45} \text{ m}^3 V=171.53π×10−45 m3V = \frac{171.5}{3} \pi \times 10^{-45} \text{ m}^3

We are given that the volume is y×10−43y \times 10^{-43} m³. So, we equate the two expressions for volume: y×10−43 m3=171.53π×10−45 m3y \times 10^{-43} \text{ m}^3 = \frac{171.5}{3} \pi \times 10^{-45} \text{ m}^3

To find yy, we rearrange the equation: y=171.53π×10−4510−43y = \frac{171.5}{3} \pi \times \frac{10^{-45}}{10^{-43}} y=171.53π×10−2y = \frac{171.5}{3} \pi \times 10^{-2}

The question asks for the value of 10y10y: 10y=10×(171.53π×10−2)10y = 10 \times \left( \frac{171.5}{3} \pi \times 10^{-2} \right) 10y=171.53π×10−110y = \frac{171.5}{3} \pi \times 10^{-1}

Now, we approximate the value using π≈3.14159\pi \approx 3.14159: 10y≈171.53×3.14159×0.110y \approx \frac{171.5}{3} \times 3.14159 \times 0.1 10y≈57.1666...×3.14159×0.110y \approx 57.1666... \times 3.14159 \times 0.1 10y≈179.59305×0.110y \approx 179.59305 \times 0.1 10y≈17.95930510y \approx 17.959305

We need to find the nearest integer to 17.95930517.959305. The nearest integer is 18.