Which one of the following statements about friction is correct?

At the angle of repose, the incline angle equals the angle of friction

The maximum force of static friction is independent of the normal reaction between the surfaces

Kinetic friction increases with the area of contact between the two surfaces, for the same normal force

The coefficient of kinetic friction is always greater than the coefficient of static friction for the same pair of surfaces

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CBSE Class 11 Science Physics · Ch 4: Laws of Motion

CBSE Class 11 Science Physics · Ch 4: Laws of Motion is an online chapter test in the CBSE Class 11 Science pattern, covering Laws of Motion. Attempt it on SeatBeat with a timer like the real exam, then see a detailed solution for every question and your score and weak-topic analysis.

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Negative marking

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Topics in this test

Topic	Subject	Questions
Laws of Motion	Physics	30

Difficulty: 7 medium · 23 hard

Laws of Motion MCQ questions with answers

The first 3 of 30 questions. Unlock the series for the rest, each with a solution.

Q1. Three concurrent coplanar forces $P$, $Q$ and $R$ keep a particle in equilibrium. The angle between $Q$ and $R$ is $90°$, and the angles between $R$ \& $P$ and between $P$ \& $Q$ are each $135°$. If $Q = R = 10\sqrt2\text{ N}$, what is the magnitude of $P$ by Lami's theorem?
तीन संगामी समतलीय (concurrent coplanar) बल $P$, $Q$ और $R$ एक कण को संतुलन में रखते हैं। $Q$ और $R$ के बीच का कोण $90°$ है, तथा $R$ व $P$ के बीच और $P$ व $Q$ के बीच प्रत्येक कोण $135°$ है। यदि $Q = R = 10\sqrt2\text{ N}$ है, तो लैमी के प्रमेय (Lami's theorem) द्वारा $P$ का परिमाण क्या है?
- A. $10\sqrt2\text{ N}$
- B. $20\text{ N}$
- C. $20\sqrt2\text{ N}$
- D. $10\text{ N}$
Solution
By Lami's theorem, $\frac{P}{\sin90°}=\frac{Q}{\sin135°}$, so $P = Q\dfrac{\sin90°}{\sin135°} = 10\sqrt2 \times \dfrac{1}{\frac{\sqrt2}{2}} = 20\text{ N}$.
लैमी के प्रमेय के अनुसार, $\frac{P}{\sin90°}=\frac{Q}{\sin135°}$, इसलिए $P = Q\dfrac{\sin90°}{\sin135°} = 10\sqrt2 \times \dfrac{1}{\frac{\sqrt2}{2}} = 20\text{ N}$।
Q2. A block is placed on the horizontal floor of a truck. For the block not to slide when the truck accelerates forward at $2.5\text{ m/s}^2$ on a level road, what is the minimum coefficient of static friction required between the block and the floor? (Take $g=10\text{ m/s}^2$.)
एक गुटका एक ट्रक की क्षैतिज फर्श पर रखा है। समतल सड़क पर जब ट्रक $2.5\text{ m/s}^2$ के त्वरण से आगे बढ़ता है, तो गुटके को फिसलने से रोकने के लिए गुटके और फर्श के बीच न्यूनतम स्थैतिक घर्षण गुणांक कितना आवश्यक है? ($g=10\text{ m/s}^2$ लें।)
- A. $4$
- B. $0.4$
- C. $0.025$
- D. $0.25$
Solution
In the truck's frame, a backward pseudo-force $ma$ on the block must be balanced by friction, so $\mu_{min} \ge \frac{a}{g} = \frac{2.5}{10}=0.25$.
ट्रक के परिप्रेक्ष्य (frame) में, गुटके पर पीछे की ओर लगने वाले आभासी बल $ma$ को घर्षण द्वारा संतुलित किया जाना चाहिए, इसलिए $\mu_{min} \ge \frac{a}{g} = \frac{2.5}{10}=0.25$।
Q3. Which one of the following statements about friction is correct?
घर्षण के बारे में निम्नलिखित में से कौन-सा कथन सही है?
- A. The maximum force of static friction is independent of the normal reaction between the surfaces / अधिकतम स्थैतिक घर्षण बल, सतहों के बीच अभिलंब प्रतिक्रिया से स्वतंत्र होता है
- B. Kinetic friction increases with the area of contact between the two surfaces, for the same normal force / समान अभिलंब बल के लिए, दो सतहों के बीच संपर्क क्षेत्रफल बढ़ने पर गतिज घर्षण बढ़ता है
- C. The coefficient of kinetic friction is always greater than the coefficient of static friction for the same pair of surfaces / समान सतहों की जोड़ी के लिए, गतिज घर्षण गुणांक सदैव स्थैतिक घर्षण गुणांक से अधिक होता है
- D. At the angle of repose, the incline angle equals the angle of friction / विश्राम कोण पर, आनत तल का कोण घर्षण कोण के बराबर होता है
Solution
At the angle of repose, the incline angle equals the angle of friction, so $\tan\theta=\mu_s$; friction instead depends on the normal reaction and not on the contact area.
विश्राम कोण पर, आनत तल का कोण घर्षण कोण के बराबर होता है, इसलिए $\tan\theta=\mu_s$; इसके अलावा घर्षण अभिलंब प्रतिक्रिया (normal reaction) पर निर्भर करता है, संपर्क क्षेत्रफल पर नहीं।

Frequently asked questions

How many questions are in the CBSE Class 11 Science Physics · Ch 4: Laws of Motion?

30 questions for 30 marks, to be answered in 20 minutes.

Is there negative marking in the CBSE Class 11 Science Physics · Ch 4: Laws of Motion?

No — wrong answers are not penalised in this test.

Which topics does the CBSE Class 11 Science Physics · Ch 4: Laws of Motion cover?

Laws of Motion (30 questions).

Is the CBSE Class 11 Science Physics · Ch 4: Laws of Motion free?

It is part of the CBSE Class 11 Science Test Series 2027 — Chapter-wise + Full Tests (43 tests, ₹199). You can first try Biology · Ch 1: The Living World for free.

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