
The mosquito net over a \[7\;{\text{ft}} \times 4\;{\text{ft}}\]bed is $3\;{\text{ft}}$ high. The net has a hole at one corner of the bed through which a mosquito enters the net. Find the magnitude of the displacement of the mosquito.
A) $\sqrt {54} \;{\text{ft}}$
B) $81\;{\text{ft}}$
C) $\sqrt {74} \;{\text{ft}}$
D) $\sqrt {64} \;{\text{ft}}$
Answer
219k+ views
Hint: The above problem is based on the kinematics. The displacement is defined as the change in the position and direction of the object. The net over the bed is the same as the box in the shape of a cube. The displacement of the mosquito can be found by the formula to find the length of the diagonal of the box.
Complete step by step answer
Given: The length of the bed is $l = 7\;{\text{ft}}$, width of the bed is $b = 4\;{\text{ft}}$ and height of the net is $h = 3\;{\text{ft}}$.
The formula to calculate the magnitude of the displacement of the mosquito is given as:
$d = \sqrt {{l^2} + {b^2} + {h^2}} $
Substitute $7\;{\text{ft}}$for l, $4\;{\text{ft}}$for b and $3\;{\text{ft}}$for h in the above formula to find the magnitude of the displacement of the mosquito.
$d = \sqrt {{{\left( {7\;{\text{ft}}} \right)}^2} + {{\left( {4\;{\text{ft}}} \right)}^2} + {{\left( {3\;{\text{ft}}} \right)}^2}} $
$d = \sqrt {74} \;{\text{ft}}$
Thus, the magnitude of the displacement of the mosquito is $\sqrt {74} \;{\text{ft}}$and the
option (A) is the correct answer.
Additional Information: The displacement and distance of the object is different. The displacement consists both magnitude and direction while distance only consists of magnitude. The change in the displacement of the object with time is the same as the velocity and change in the distance of the object with time is the same as the speed of the object.
Note: The displacement of the mosquito can also be found by using the vector method. The one corner of the net assumed as the origin and coordinates as (0, 0, 0) and the coordinates of the mosquito as (7 ft, 4 ft, 3 ft). Then find the distance between these two coordinates to find the magnitude of the mosquito.
Complete step by step answer
Given: The length of the bed is $l = 7\;{\text{ft}}$, width of the bed is $b = 4\;{\text{ft}}$ and height of the net is $h = 3\;{\text{ft}}$.
The formula to calculate the magnitude of the displacement of the mosquito is given as:
$d = \sqrt {{l^2} + {b^2} + {h^2}} $
Substitute $7\;{\text{ft}}$for l, $4\;{\text{ft}}$for b and $3\;{\text{ft}}$for h in the above formula to find the magnitude of the displacement of the mosquito.
$d = \sqrt {{{\left( {7\;{\text{ft}}} \right)}^2} + {{\left( {4\;{\text{ft}}} \right)}^2} + {{\left( {3\;{\text{ft}}} \right)}^2}} $
$d = \sqrt {74} \;{\text{ft}}$
Thus, the magnitude of the displacement of the mosquito is $\sqrt {74} \;{\text{ft}}$and the
option (A) is the correct answer.
Additional Information: The displacement and distance of the object is different. The displacement consists both magnitude and direction while distance only consists of magnitude. The change in the displacement of the object with time is the same as the velocity and change in the distance of the object with time is the same as the speed of the object.
Note: The displacement of the mosquito can also be found by using the vector method. The one corner of the net assumed as the origin and coordinates as (0, 0, 0) and the coordinates of the mosquito as (7 ft, 4 ft, 3 ft). Then find the distance between these two coordinates to find the magnitude of the mosquito.
Recently Updated Pages
Two discs which are rotating about their respective class 11 physics JEE_Main

A ladder rests against a frictionless vertical wall class 11 physics JEE_Main

Two simple pendulums of lengths 1 m and 16 m respectively class 11 physics JEE_Main

The slopes of isothermal and adiabatic curves are related class 11 physics JEE_Main

A trolly falling freely on an inclined plane as shown class 11 physics JEE_Main

The masses M1 and M2M2 M1 are released from rest Using class 11 physics JEE_Main

Trending doubts
JEE Main 2026: Application Form Open, Exam Dates, Syllabus, Eligibility & Question Papers

Understanding Uniform Acceleration in Physics

Derivation of Equation of Trajectory Explained for Students

Hybridisation in Chemistry – Concept, Types & Applications

Understanding the Angle of Deviation in a Prism

Understanding Collisions: Types and Examples for Students

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

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

NCERT Solutions For Class 11 Physics Chapter 8 Mechanical Properties Of Solids

Motion in a Straight Line Class 11 Physics Chapter 2 CBSE Notes - 2025-26

NCERT Solutions for Class 11 Physics Chapter 7 Gravitation 2025-26

Understanding Atomic Structure for Beginners

