A block of wood is kept on the table top. The mass of wooden block is $5 \mathrm{kg}$ and its dimensions are $40 \mathrm{cm} \times 20 \mathrm{cm} \times 10 \mathrm{cm} .$ Find the pressure exerted by the wooden
block on the table top, if it is made to lie on the table top with its sides of dimensions (a) $20 \mathrm{cm} \times 10 \mathrm{cm}$ and $(\mathrm{b}) 40 \mathrm{cm} \times 20 \mathrm{cm}$
A (a)$P=2440 N m^{-2}$ (b) $P=610.5 N m^{-2}$
B (a)$P=2450 \mathrm{Nm}^{-2}$ (b) $\mathrm{P}=612.5 \mathrm{Nm}^{-2}$
C (a)$P=2460 N m^{-2}$ (b) $P=618.5 N m^{-2}$
D (a)$P=2470 N m^{-2}$ (b) $P=712.5 N m^{-2}$
Answer
258k+ views
Hint: We should know that velocity is defined as the rate change of displacement per unit time. Speed in a specific direction is also known as velocity. Velocity is equal to displacement divided by time. Speed, being a scalar quantity, is the rate at which an object covers distance. The average speed is the distance which is a scalar quantity per time ratio. On the other hand, velocity is a vector quantity; it is direction-aware. An object which moves in the negative direction has a negative velocity. If the object is slowing down then its acceleration vector is directed in the opposite direction as its motion in this case. Based on this we have to solve this question.
Complete step by step answer
From the data given in the question,
We see that,
Mass of the wooden block, $\mathrm{m}=5 \mathrm{kg}$
Thrust due to wooden block, $\mathrm{Mg}=5 \mathrm{kg} \times 9.8 \mathrm{ms}^{-2}=49 \mathrm{N}$
(a) Surface area of $20 \mathrm{cm} \times 10 \mathrm{cm}$ surface,
$\mathrm{A}=(20 \times 10) \mathrm{cm}^{2}$
$=(0.2 \times 0.1) \mathrm{m}^{2}$
$=0.02 \mathrm{m}^{2}$
$P=\dfrac{m g}{A}$
With the given equation, we can derive,
$\mathrm{P}=\dfrac{49 \mathrm{N}}{0.02 \mathrm{m}^{2}}=\dfrac{49}{0.02} \mathrm{Nm}^{-2}$
or Pressure, $\mathrm{P}=2450 \mathrm{Nm}^{-2}$.
(b) Surface area of $40 \mathrm{cm} \times 20 \mathrm{cm}$ surface,
$\mathrm{A}=(40 \times 20) \mathrm{cm}^{2}$
$=(0.4 \times 0.2) \mathrm{m}^{2}=0.08 \mathrm{m}^{2}$
Equating from the derived data, we get,
$\mathrm{P}=\dfrac{49 \mathrm{N}}{0.08 \mathrm{m}^{2}}=\dfrac{49}{0.08} \mathrm{Nm}^{-2}$
or Pressure, $P=612.5 \mathrm{Nm}^{-2}$
Therefore, the correct answer is Option B.
Note: We should know that if an object's speed or velocity is increasing at a constant rate then we say it has uniform acceleration. The rate of acceleration is constant. If a car speeds up then slows down then speeds up it doesn't have uniform acceleration. The instantaneous acceleration, or simply acceleration, is defined as the limit of the average acceleration when the interval of time considered approaches 0. It is also defined in a similar manner as the derivative of velocity with respect to time. If an object begins acceleration from rest or a standstill, its initial time is 0. If we get a negative value for acceleration, it means the object is slowing down. The acceleration of an object is its change in velocity over an increment of time. This can mean a change in the object's speed or direction. Average acceleration is the change of velocity over a period of time. Constant or uniform acceleration is when the velocity changes the same amount in every equal time period.
Complete step by step answer
From the data given in the question,
We see that,
Mass of the wooden block, $\mathrm{m}=5 \mathrm{kg}$
Thrust due to wooden block, $\mathrm{Mg}=5 \mathrm{kg} \times 9.8 \mathrm{ms}^{-2}=49 \mathrm{N}$
(a) Surface area of $20 \mathrm{cm} \times 10 \mathrm{cm}$ surface,
$\mathrm{A}=(20 \times 10) \mathrm{cm}^{2}$
$=(0.2 \times 0.1) \mathrm{m}^{2}$
$=0.02 \mathrm{m}^{2}$
$P=\dfrac{m g}{A}$
With the given equation, we can derive,
$\mathrm{P}=\dfrac{49 \mathrm{N}}{0.02 \mathrm{m}^{2}}=\dfrac{49}{0.02} \mathrm{Nm}^{-2}$
or Pressure, $\mathrm{P}=2450 \mathrm{Nm}^{-2}$.
(b) Surface area of $40 \mathrm{cm} \times 20 \mathrm{cm}$ surface,
$\mathrm{A}=(40 \times 20) \mathrm{cm}^{2}$
$=(0.4 \times 0.2) \mathrm{m}^{2}=0.08 \mathrm{m}^{2}$
Equating from the derived data, we get,
$\mathrm{P}=\dfrac{49 \mathrm{N}}{0.08 \mathrm{m}^{2}}=\dfrac{49}{0.08} \mathrm{Nm}^{-2}$
or Pressure, $P=612.5 \mathrm{Nm}^{-2}$
Therefore, the correct answer is Option B.
Note: We should know that if an object's speed or velocity is increasing at a constant rate then we say it has uniform acceleration. The rate of acceleration is constant. If a car speeds up then slows down then speeds up it doesn't have uniform acceleration. The instantaneous acceleration, or simply acceleration, is defined as the limit of the average acceleration when the interval of time considered approaches 0. It is also defined in a similar manner as the derivative of velocity with respect to time. If an object begins acceleration from rest or a standstill, its initial time is 0. If we get a negative value for acceleration, it means the object is slowing down. The acceleration of an object is its change in velocity over an increment of time. This can mean a change in the object's speed or direction. Average acceleration is the change of velocity over a period of time. Constant or uniform acceleration is when the velocity changes the same amount in every equal time period.
Recently Updated Pages
Circuit Switching vs Packet Switching: Key Differences Explained

Dimensions of Pressure in Physics: Formula, Derivation & SI Unit

JEE General Topics in Chemistry Important Concepts and Tips

JEE Extractive Metallurgy Important Concepts and Tips for Exam Preparation

JEE Atomic Structure and Chemical Bonding important Concepts and Tips

JEE Amino Acids and Peptides Important Concepts and Tips for Exam Preparation

Trending doubts
JEE Main 2026: Exam Dates, Session 2 Updates, City Slip, Admit Card & Latest News

JEE Main Participating Colleges 2026 - A Complete List of Top Colleges

JEE Main 2026 Application Login: Direct Link, Registration, Form Fill, and Steps

JEE Main Colleges 2026: Complete List of Participating Institutes

JEE Main Marking Scheme 2026- Paper-Wise Marks Distribution and Negative Marking Details

Hybridisation in Chemistry – Concept, Types & Applications

Other Pages
JEE Advanced 2026 - Exam Date (Released), Syllabus, Registration, Eligibility, Preparation, and More

JEE Advanced Marks vs Ranks 2025: Understanding Category-wise Qualifying Marks and Previous Year Cut-offs

CBSE Notes Class 11 Physics Chapter 1 - Units And Measurements - 2025-26

NCERT Solutions For Class 11 Physics Chapter 1 Units And Measurements - 2025-26

Important Questions For Class 11 Physics Chapter 1 Units and Measurement - 2025-26

JEE Advanced Weightage 2025 Chapter-Wise for Physics, Maths and Chemistry

